

















































































Class_ l L 5 40 

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DAMS AND WEIRS 


AN ANALYTICAL AND PRACTICAL TREATISE ON GRAVITY 
DAMS AND WEIRS; ARCH AND BUTTRESS DAMS; 
SUBMERGED WEIRS; AND BARRAGES 


BY 

W. G. BLIGH 

n 

FORMERLY EXECUTIVE ENGINEER, PUBLIC WORKS DEPARTMENT, INDIA, AND 
INSPECTING ENGINEER, DEPARTMENT OF INTERIOR, CANADA 
MEMBER OF THE INSTITUTION OF CIVIL ENGINEERS 
MEMBER OF AMERICAN SOCIETY OF CIVIL ENGINEERS 
MEMBER OF CANADIAN SOCIETY OF CIVIL ENGINEERS 


ILLUSTRATED 



AMERICAN TECHNICAL SOCIETY 
CHICAGO 






COPYRIGHT, 1916, 1917, BY 
AMERICAN TECHNICAL SOCIETY 


COPYRIGHTED IN GREAT BRITAIN 
ALL RIGHTS RESERVED 

* 



FEB 23 1918 



©CI.A 481795 





INTRODUCTION 


AN unused waterfall, no matter how beautiful, appeals to an 
engineer mainly as an economic waste, and he fairly aches 
to throw a dam across the rushing torrents or to utilize 
the power of the water which glides gracefully over the falls and 
dashes itself into useless spray many feet below. His progress in 
the past years, however, has in no way measured up to his desires, 
but with the United States and other governments undertaking 
gigantic irrigation projects in order to reclaim vast areas of till¬ 
able lands and with hydroelectric companies acquiring the power 
rights of our great waterfalls, the last few years have witnessed 
wonderful progress in this type of engineering work. The use of 
reinforced concrete as a standard material and the solving of the 
many problems in connection with it has greatly simplified and 
cheapened the construction, thus avoiding the greater difficulties 
of masonry construction usually found in the older dams. 

All of this progress in the design of dams and weirs, however, 
has served to multiply the types of design and has increased 
the need for an authoritative and up-to-date treatise on the 
theoretical and practical questions involved. The author of this 
work has been a designing engineer for more than a generation 
and has built dams and weirs in India and this country, besides 
inspecting irrigation works in Canada. He is, therefore, abun¬ 
dantly qualified to speak not only from the historic side of the 
work but from the modern practical side as well. In addition 
to a careful analysis of each different type of profile, he has given 
critical studies of the examples of this type, showing the good 
and bad points of the designs. A wealth of practical prob¬ 
lems together with their solution makes the treatise exceedingly 
valuable. 

<J It is the hope of the publishers that this modern treatise 
will satisfy the demand for a brief but authoritative work on 
the subject and that it will find a real place in the technical 
literature of Civil Engineering. 






CONTENTS 


PAGE 

Gravity dams. 2 

Pressure of water on wall. 2 

Method for graphical calculations. 2 

Conditions of “middle third” and limiting stress. 3 

Compressive stress limit. 4 

Design of dams. 4 

Theoretical profile. 4 

Practical profile. 8 

Crest width. 9 

Rear widening. 10 

Variation of height.. 13 

High and wide crest. 13 

Graphical method. 16 

Analytical, method. 18 

Pressure distribution. 23 

Graphical method for distribution of pressure. 25 

Maximum pressure limit. 27 

Limiting height. 29 

Internal shear and tension. 30 

Security against failure by sliding or shear. 31 

Influence lines. 31 

Actual pressures in figures. 34 

Haessler’s method. 36 

Stepped polygon. 37 

Modified equivalent pressure area in inclined back dam. 37 

Curved back profiles. 39 

Treatment for broken line profiles. 41 

Example of Haessler’s method. 42 

Relations of R. N. and W. 43 

Unusually high dams. 43 

Pentagonal profile to be widened. 47 

Silt against base of dam. 50 

Ice pressure. 51 

Partial overfall dams. % . 52 

Notable existing dams. 53 

Cheeseman Lake dam. 53 

Analytical check. 55 

Roosevelt dam. 56 

New Croton dam. 58 

Assuan dam.. . . .. 59 

Cross River and Ashokan darns. 65 

Burrin Juick dam. 65 

Arrow Rock dam. 67 

Special foundations. 69 

Aprons affect uplift. 70 















































CONTENTS 


Special foundations—(continued) page 

Rear aprons decrease uplift. 71 

Rock below gravel. 72 

Gravity dam reinforced against ice pressure. 73 

Gravity overfall dams or weirs. 75 

Characteristics of overfalls. 75 

Approximate crest width. 77 

Pressures affected by varying water level. 79 

Method of calculating depth of overfall. 82 

Obj ections to ‘ 1 Ogee ’ ’ overfalls. 85 

Folsam weir. 85 

Dhukwa weir. 90 

Mariquina weir. 92 

Granite Reef weir. 92 

Nira weir. 95 

Castlewood weir. 96 

American dams on pervious foundations. 97 

Arched dams. 101 

» 

Theoretical and practical profiles. 102 

Support of vertical water loads in arched dams. 104 

Pathfinder dam. 104 

Shoshone dam. 107 

Sweetwater dam. 109 

Barossa dam. Ill 

Burrin Juick subsidiary dam. 112 

Dams with variable radii. 112 

Multiple arch or hollow arch buttress dams. 113 

Multiple arch generally more useful than single arch dams. 113 

Mir Alam dam. 114 

Stresses in buttress. 117 

Belubula dam. 118 

Ogden dam.... . 120 

Design for multiple arch dam. 122 

Reverse water pressure. 124 

Pressure on foundations. 125 

Flood pressures. 129 

Big Bear Valley dam. 131 

Hollow slab buttress dams. 136 

Formulas for reinforced concrete. 137 

Guayabal dam. 141 

Bassano dam.1. 146 

Submerged weirs founded on sand. 151 

Percolation beneath dam. 152 

Governing factor for stability.... 153 

Vertical obstruction to percolation. 159 

Rear apron. 159 

Example of design, type A. 164 

Discussion of Narora weir. 167 

Sloping apron weirs, type B. 169 


















































CONTENTS 


Submerged weirs founded on sand—(continued) page 

Restoration of Khanki weir. 171 

Merala weir. 171 

Porous fore aprons. 173 

Okhla and Madaya weirs. 177 

Dehri weir... 178 

Laguna weir. 179 

Damietta and Rosetta weirs. 179 

Open dams or barrages.... 182 

Barrage defined. 182 

Weir sluices of Corbett dam. 189 

General features of river regulators. 193 

Stability of Assiut barrage. 194 

Hindia barrage. 194 

American vs. Indian treatment. 196 

North Mon canal. 200 

Upper Coleroon regulator. 201 

St. Andrew’s Rapids dam. 201 

Automatic dam or regulator. 205 





















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DAMS AND WEIRS 


PART 1 


INTRODUCTION 

1. Definitions. A dam may be defined as an impervious wall 
of masonry, concrete, earth, or loose rock which upholds a mass of 
water at its rear, while its face or lower side is free from the pressure 
of water to any appreciable extent. The waste water of the reser¬ 
voir impounded by the dam is disposed of by means of a waste weir, 
or by a spillway clear of the work, or in rare cases, by sluice openings 
in the body of the dam. 

Weirs, or overfall dams, although often confounded with bulk¬ 
head dams, differ from the latter in the following points, first, 
that the water overflows the crest, and second, that the tail water 
is formed below the dam. These two differences often modify the 
conditions of stress which are applicable in the design of dams 
proper, and consequently the subject of weirs demands separate 
treatment. 

2. Classification. Dams and weirs may be classified as 
follows: 

1. Gravity Dams 

2. Gravity Overfalls, or Weirs 

3. Arched Dams 

4. Hollow Arch Buttress Dams 

5. Hollow Slab Buttress Dams 

6. Submerged Weirs 

7. Open Darns, or Barrages 

The subjects of earth, rock fill, and steel dams will not be 
treated in this article, as the matter has been already dealt with in 
other volumes. Graphical as well as analytical methods will be 
made use of, the former procedure being explained in detail as 
occasion demands. 



2 


DAMS AND WEIRS 


GRAVITY DAMS 


GENERAL DISCUSSION OF DAMS 


A gravity dam is one in which the pressure of the water is 
resisted by the weight or “gravity” of the dam itself. 

3. Pressure of Water on Wall. The hydrostatic pressure of 
the water impounded by a wall or dam may be graphically repre¬ 
sented by the area of a triangle with its apex at the surface and its 
base drawn normal to the back line of the dam, which base is equal 
or proportionate to the vertical depth. When the back of the wall 

H 2 

is vertical, as in Fig. 1, the area of this pressure triangle will be - 


H being the vertical height. When, as in Fig. 2, the back is inclined, 


this area will be 


H' being the inclined length of the exposed 


surface, which equals II cosec </>. 

The actual pressure of water per unit length of dam is the 
above area multiplied by the unit weight of water. This unit 



Fig. 1. Water Pressure Area 
with Back of Dam Vertical 



Fig. 2. Water Pressure Area with 
Back of Dam Inclined 


weight is symbolized by w, which is 62.5 pounds, or ^ short ton, or 
sV long ton, per cubic foot. 

Unit Pressure. The pressure per square foot, or unit pressure 
on the wall at any point, is measured by the corresponding ordinate 
of the above triangle, drawn parallel to its base, and is evidently 
the same in both Figs. 1 and 2. The total pressure on the inclined 
back as represented by the triangle in Fig. 2 will, however, be 
greater than in Fig. 1. 

4. Method for Graphical Calculations. For graphical calcula¬ 
tions when forces of dissimilar unit weight or specific gravity are 







































DAMS AND WEIRS 


3 


engaged, as in the case of water and masonry, or earth and masonry, 
it is the usual practice to reduce them to one common denominator 
by making alterations in the areas of one or the other, the weight of 
the masonry being usually taken as a standard. This result is 
effected by making the bases of the triangles of water pressure 

H 

equal, not to H, but to —, p (rho) being the sign of the specific 

P 

gravity of the solid material in the wall. The triangle thus reduced 
will then represent a weight or area of masonry 1 unit thick, equiva¬ 
lent to that of water. This device enables the item of unit weight, 
which is wXp to be eliminated as a common factor from the forces 
engaged, i. e., of the water pressure and of the weight of the wall. 
The factor thus omitted has to be multiplied in again at the close 
of the graphical operation, only, however, in cases where actual 
pressures in tons or pounds are required to be known. 

Value of p. The values ordinarily adopted for p, the specific 
gravity of masonry or concrete, are 2\ and 2.4, i.e., equivalent to 
weights of 141 and 150 pounds, respectively, per cubic foot, while 
for brickwork 2 is a sufficiently large value. The value of wp in the 

3 

former case will be .069 ton and in the latter .075, or — ton. 

In some cases the actual value of p mounts as high as 2.5 and 
even 2.7, when heavy granite or basalt is the material employed. 

The reduction thus made in the water pressure areas has further 
the convenience of reducing the space occupied by the diagram. 
The areas of the reduced triangles of water pressure in Figs. 1 and 
H 2 H'H 

2 are —— and —-— respectively. 

2 p Ip 

5. Conditions of “Middle Third” and Limiting Stress. Sec¬ 
tions of gravity dams are designed on the well-known principle of 
the “middle third.” This expression signifies that the profile of 
the wall must be such that the resultant pressure lines or centers of 
pressure due first to the weight of the dam considered alone, and 
second with the external water pressure in addition, must both 
fall within the middle third of the section on any horizontal base. 
These two conditions of stress are designated, Reservoir Empty 
(R.E.) and Reservoir Full (R.F.). The fulfillment of this condition 
insures the following requirement: The maximum compressive ver- 




4 


DAMS AND WEIRS 


tical unit stress ( s ), or reaction on the base of a dam, shall not exceed 
twice the mean compressive unit stress, or, stated symbolically, 

s^2si 

Now the mean vertical compressive unit stress Si is the weight of the 
structure divided by its base length- i.e., 

w 
sv =7 

2JV 

Hence, s, the maximum vertical unit pressure, should not exceed —— 

Further comments on the distribution of the reaction on the base 
of a dam will be made in a later paragraph. 

6. Compressive Stress Limit. A second condition imposed is 
that of the internal compressive stress limit, that is: The maximum 
permissible compressive unit stress which is developed in the interior 
of the masonry of the dam, must not be exceeded. This value can be 
experimentally found by crushing a cube of the material employed, 
and using a factor of safety of 6 or 8. Cement concrete will crush 
at about 2000 pounds per square inch, equivalent to 144 tons (of 
2000 pounds) per square foot. The safe value of s would then be 
144 

—- = 18 tons per square foot. For ordinary lime concrete as 
8 

employed in the East, the limit pressure adopted is generally 8 
“long” tons, equivalent to 9 tons of 2000 pounds. Ten “long” 
tons, or 11.2 “short” tons is also a common value. 


DESIGN OF DAMS 


7. Theoretical Profile. The theoretically correct profile of a 
so-termed “low” masonry dam, i.e., one of such height that the limit 
stress is not attained under the conditions above outlined, is that 
of a right-angled triangle having its back toward the water vertical, 
and its apex at the water surface. It can be proved that the proper 
base width b of this triangle is expressed by the formula 



This profile, shown in Fig. 3, will be termed the “elementary trian¬ 
gular profile”, as on it the design of all profiles of dams is more or 
less based. In this expression, II is the vertical height. The base 





DAMS AND WEIRS 


5 


width of 



insures the exact incidence of the vertical resultant (IF) 


(R.E.) and of the inclined resultant R (R.F.) at the inner and outer 
edge, respectively, of the central third division of the base. The 
condition of the middle third is thus fulfilled in the most economical 
manner possible, a factor of safety of 2 against overturning is 
obtained, and further, the angle of inclination of the resultant R 
with regard to the base is usually such as to preclude danger of 
failure by sliding. 

The fore slope or hypothenuse will be in the ratio 1: ^p which, 
when p = 2|, will equal 2:3, a slope very commonly adopted, 


F K_ i p 




Fig. 3. Elementary Triangular Profile for “Low” Masonry Dam 


and with p = 2.4 the ratio will be 1:1.549. The area of the ele- 

H 2 

mentary triangle is — j=- while, as we have seen, that of the water 

2 V p 

II 2 

pressure is -—. 6 is the vertical angle between II and R, and 

2 p 

sec 6 = = 1.187 with p = 2.4. 

Vp 

In Fig. 3 the resultant pressure lines are drawn to intersect the 
base so as to afford ocular proof of the stability of the section under 
the postulated conditions. 

Graphical Method. The graphical procedure will now be briefly 
explained, and also in the future as fresh developments arise, for 













































6 


DAMS AND WEIRS 


the benefit of those who are imperfectly acquainted with this valu¬ 
able labor-saving method. 

There are two forces engaged, P the horizontal, or, it may be Pi, 
the inclined water pressure acting through the center of gravity of its 
area normal to the back of the wall, and W the weight or area of the 
wall. Of these two forces the item wp, or unit weight, has already 
been eliminated as a common factor, leaving the pressures repre¬ 
sented by superficial areas. As, however, the height H is also com¬ 
mon to both triangles, this can likewise be eliminated. The forces 
may then be represented simply by the half widths of the triangular 
areas by which means all figuring and scaling may be avoided. 

First, a force polygon has to be constructed. In Fig. 3a, P is 
first drawn horizontally to designate the water pressure, its length 
being made equal to the half width of its pressure area in Fig. 3. 
From the extremity of P, the load line W is drawn vertically, equal 
to the half width of the elementary triangular profile, then the 
closing line R according to the law of the triangle of forces will 
represent the resultant in magnitude and direction. Second, the 
lines of actual pressures reciprocal to those on the force polygon 
will have to be transferred to the profile. The incidence of the 
resultant water pressure on the back is that of a line drawn through 

the c.g. of the area of pressure, parallel to its base, in this case, at —, 

or one-third the height of the water-pressure triangle, above the 
base. Its direction, like that of the base, is normal to the back, in 
this case horizontal, and if prolonged it will intersect the vertical 
force IF, which in like manner acts through the center of gravity of 
the elementary profile of the wall. From this point of intersection 
the resultant R is drawn parallel to its reciprocal in Fig. 3a. Both 
IF and R are continued until they cut the base line, and these points 
of intersection will be found to be exactly at the inner and outer 
edges of the middle third division of the base. It will be seen that 
when the reservoir is empty the center of pressure on the base is at 
the incidence of IF, when full it is shifted to that of R. 

Analytical Method. The same proof can be made analytically 
as follows: The weight of the two triangles IF and P can be repre- 

H H 

sented by their bases which are —= and —, respectively. If moments 

Vp p 



DAMS AND WEIRS 


7 


be taken about the outer edge of the middle third, the lever arm of 

b H 

the vertical force 11 is clearly — or 7 — 7 = and that of P, the horizontal 

O O V p 

force, is the distance of the center of gravity of the triangle of water 
pressure above the base, viz, The equation will then stand 

O 


( 


II 

r=x 


Vp 3 V/ 


iSHfxD- 


or 


II 2 II 2 


= 0 


3 p 3 p 

If the actual values of R and of W were required, their measured or 
calculated lengths would have to be multiplied by //and by wp in order 
to convert them to tons, pounds, or kilograms, as may be required. 
In many, in fact most, cases actual pressures are not required to be 
known, only the position of the centers of pressures in the profile. 

Thus a line of pressures can be traced through a profile giving 
the positions of the centers of pressure without the necessity of 
converting the measured lengths into actual quantities. In the 
elementary triangle, Fig. 3, the value of the vertical resultant W is 

II w-ylp. 0 f r required in the older methods of calculation 


2 


is 


// 2 wVp+1 




The following values relative to p will be found useful. 


P OR 
Specific 
Gravity 

Vp 

1 

Vp 

, i_ 

P 

Pounds per 
Cubic Foot 

Tons per 
Cubic Foot 

2 

1.414 

.71 

.5 

125 

.0625 

21 

1.5 

2 

3 

4 

9 

141 

.07 

2.4 

1.55 

.645 

.417 

150 

. 075 = 4 3 q 

2.5 

1.58 

.633 

.4 

156 

.078 

2.7 

1.643 

.609 

.37 

168.7 

.084 


Profile with Back Inclined. If the elementary profile be canted 
forward so that its back is inclined to the vertical, it will be found 
that the incidence of R will fall outside the middle third while that 
of W will be inside. The base will, therefore, have to be increased 

above - 7 =-. 

v p 























8 


DAMS AND WEIRS 


When the back is overhanging, on the other hand, R will fall 
inside and W outside the middle third. The vertically backed 
section is consequently the most economical. 

8 . Practical Profile. In actual practice a dam profile must be 
provided with a crest of definite width, and not terminate in the 
apex of a triangle. The upper part of a dam is subjected to indefi¬ 
nite but considerable stresses of an abnormal character, due to extreme 
changes in temperature, consequently a solid crest is a necessity. 
The imposition of a rectangular crest, as shown in dotted lines on 



Fig. 3, transforms the triangular profile into a pentagon. This has 
the effect of increasing the stability of the section (R.F.) so that 
the base width can be somewhat reduced, at the same time the 
vertical resultant IF (R.E.), falls outside the middle third, but to 
so small an extent that this infringement of the imposed condition 
is often entirely neglected. In order to provide against this, a strip 
of material will have to be added to the back of the plain pentagonal 
profile. Fig. 4 is a diagram explanatory of these modifications. 
The dimensions of this added strip, as well as its position, can be 


















DAMS AND WEIRS 


9 


conveniently expressed in terms of (k) the crest width—i.e., AB 
in Fig. 4. The line of pressure (R.E.) will begin to leave the middle 
third at the depth AD, which is found by calculation which need 

not be produced here, to be 2Wp. Below the point D, the 
divergence of the line of pressure will continue for a further depth 

DE, the point E, being close upon 3.1AAp below the crest, or 

l.l&Vp below D. Below point E, the line of pressure will no 
longer diverge outward, but will tend to regain its original position, 
consequently no further widening will be necessary, and the added 



strip will be rectangular in form down to the base. The points D 
and F being joined, this portion of the back will be battered. r Ihe 
width of this added strip EF will be, with close approximation, 

— or .06&. 

16 

9. Crest Width. The crest width of a dam should be propor¬ 
tioned to its actual height in case of a “low” dam, and in the case of 
a “high” dam to the limiting height—i.e., to that depth measured 
below the crest at which the maximum stress in the masonry is first 























































10 


DAMS AND WEIRS 


reached. Thus in “high” dams the upper part can always be of 
the same dimensions except where the requirements of cross com¬ 
munication necessitate a wider crest. 

The effect of an abnormally wide crest can be modified by 
causing it to overhang the fore slope, this widening being carried 
by piers and arches. A good example of this construction occurs 
in the Chartrain dam, Fig. 5. The arches form a stiff but light 
finish to the dam and have a pleasing architectural effect. The 
same procedure, but in a less pronounced degree, is carried out in 

the Croton dam, Fig. 27. 

The formula for crest 
width can be expressed 
either in terms of the limit¬ 
ing height Hi, or of the 
base b, where the limiting 
height is not attained, and 
a good proportion is given 
by the following empirical 
rule: 

k='iir l ( 2 ) 

or 

k = .15b (2a) 



Fig. 6. Pentagonal Profile—Back Vertical 


This latter formula makes 
the crest width a function of the specific gravity as well as of the 
height, which is theoretically sound. 

10. Rear Widening. Where the rear widening of a “low” 
dam is neglected or where a uniform batter is substituted for the 
arrangement shown in Fig. 4, the profile will be pentagonal in out¬ 
line. When the back is vertical the two triangles composing the 
body of the dam are similar. If the ratio existing between the crest 

k 

(k) and the base (b), or — be designated by r, then k = br, and h, the 


depth of the vertical side in Fig. 6, = Hr and k X h = Hbr 2 . 

In order to find what value the base width b should have, so 
that the center of pressure (R.F.) will fall exactly at the edge of the 
middle third, the moments of all the forces engaged will have to be 
taken about this point and equated to zero. The vertical forces 



















DAMS AND WEIRS 


11 


consist of W y the lower, and W\ the upper triangle; the horizontal 
of P, the water pressure. 

11. Method of Calculation. The pressures can be represented 
by the areas of the prisms involved, the triangle of water pressure 
being as usual reduced by dividing its base by p. A further elim¬ 
ination of common factors can be achieved by discarding A which 

Li 

is common to all three forces, the area TIh being represented by br 2 


because the actual original value is 


Hbr 2 


The forces then are W, 


II 


represented by b; Wi by br 2 ; and P by —; the actual value of the 

II 2 

latter being ——. The lever arm distances of the c.g.’s of these three 
2 p 

forces from A, the incidence of R, are as follows: of W, —, of Wiy 

2 H 1 

— (b — br), and of P, —. The equation will then stand, eliminating 

bxb+br 2 (2b — 2Xbr) —— = 0 


or 


whence 


b 2 (l+2r 2 -2r 3 )- —= 0 




Vp Vl+2r 2 —2r 3 


( 3 ) 


The value of b thus obtained will prove a useful guide in deciding 
the base width even when the back of the wall is not vertical, as only 
a small increase will be needed to allow for the altered profile. When 


A or r = .15 the reducing coefficient works out to * -, the reciprocal 
^ l.uiy 

of which is .981. Thus with a profile 80 feet high with p = 2.5 and 

80 

r = .15, the base width of the pentagonal profile will be b = —= X .981 


= 49.64 feet; the decrease in base width below that of the elementary 
profile without crest will be 50.60 — 49.64 = 0.96 feet. The crest 
width will be 49.64X.15 = 7.45 feet. In actual practice, the dimen¬ 
sions would be in round numbers, 50 feet base and (% feet crest 











12 


DAMS AND WEIRS 


width as made on Fig. 6 . The face of the profile in Fig. G is made 
by joining the toe of the base with the apex of the triangle of water 
pressure. 

Graphical Process . The graphical processes of finding the 
incidences of W and of R on the base are self-explanatory and are 
shown on Fig. 6 . The profile is divided into two triangular areas, 
( 1 ), 45 square feet and ( 2 ), 2000 square feet. The two final result¬ 
ants fall almost exactly at the middle third boundaries, IF, as 
might be conjectured, a trifle outside. Areas are taken instead of 
\ widths, owing to II not being a common factor. 

Analytical Process. The analytical process of taking moments 


about the heel is 

shown below 




Area 

Lever Arm 

Moment 

(i) 

45 

7.5X2 

3 

225 

(2) 

2000 

50 

3 

33333 

w 

2045 


33558 


The value 33,558, which is the total moment of parts equals the 
moment of the whole about the same point or 

2045 X.r = 33558 


x = 16.41 feet 
50 

The incidence of IF is therefore — —16.41 = .26 ft. outside the middle 

O 

third. To find that of R relative to the heel, the distance (see 

17 x , . TT r , D . PII 1280X80 102400 ir _ n 

section 17) between II and R is 7^77 = —— 7 —— = = 16.69. 

311 3X204O 613o 

The distance of R from the heel is therefore 16.69+16.41 =33.10 ft. 
2 

The — point is 33.33 feet distant, consequently the incidence of R 

O 

is .23 foot within the middle third. 

If the base and crest had been made of the exact dimensions 
deduced from the formula, the incidence of I\ would be exactly at 
2 

the — point while IF would fall slightly outside the one-third point. 

O 

















DAMS AND WEIRS 


13 


12. Variation of Height. The height of a dam is seldom uni- 

i 

form throughout; it must vary with the irregularities of the river 
bed, so that the maximum section extends for a short length only, 
while the remainder is of varying height. This situation will affect 
the relationship between the crest width and the height, and also 
the base width. To be consistent, the former should vary in width 
in proportion to the height. This, however, is hardly practicable, 
consequently the width of the crest should be based more on the 
average than on the maximum height, and could be made, wider 
wherever a dip occurs in the foundation level. 

13. High and Wide Crest. In case of a very high as well as 
wide crest, i.e., one carried much 
higher than the apex of the trian¬ 
gle of water pressure, it is not 
desirable to reduce the base width 
much below that of the elemen¬ 
tary triangle. The excess of 
material in the upper quarter of 

low” dam can be reduced by 


7 ^ 


a 


manipulating the fore slope. 

This latter, which is drawn up¬ 
ward from the toe of the base, 
in Fig. 7, can be aligned in three 
directions. First, by a line ter¬ 
minating at the apex of the trian¬ 
gle of water pressure; second, it 
can be made parallel to that of 
the elementary profile, that is, it can be given an inclination of 




C B OR C,B, = 50 . 0 ' 


CC, 


Profile Showing Different Disposition 
of Fore Slope 


_JL to the vertical, and third, the slope or batter can be made 

V p 

flatter than the last. This latter disposition is only suitable w ith 
an abnormally high and wide crest and is practically carried out 
in the Chartrain dam, Fig. 5, where the base is not reduced at all 


II 

below —j=. 

Vp 

Reduction to any large extent, of the neck of the profile thus 
effected is, however, not to be commended, as the upper quarter of 
a dam is exposed to severe though indeterminate stresses due to 














14 


DAMS AND WEIRS 


changes of temperature, wind pressure, etc., and also probably to 
masses of ice put in motion by the wind. The Cross River dam, 
to be illustrated later, as well as the Ashokan dam, are examples of 
an abnormally thick upper quarter being provided on account of 
ice. Whatever disposition of the fore slope is adopted, the profile 
should be tested graphically or analytically, the line of pressure, if 
necessary, being drawn through the profile, as will later be explained. 

From the above remarks it will be gathered that the design of 
the section of a dam down to the limiting depth can be drawn by a 
few lines based on the elementary profile which, if necessary, can 
be modified by applying the test of ascertaining the exact position 
of the centers of pressure on the base. If the incidence of these 
resultants falls at or close within the edge of the middle third divi¬ 
sion of the base, the section can be pronounced satisfactory; if 
otherwise, it can easily be altered to produce the desired result. 

Freeboard. The crest has to be raised above actual full reservoir 
level by an extent equal to the calculated depth of water passing 
over the waste weir or through the spillway, as the case may be. 
This extra freeboard, which adds considerably to the cost of a 
work, particularly when the dam is of great length and connected 
with long embankments, can be avoided by the adoption of auto¬ 
matic waste gates by which means full reservoir level and high flood 
level are merged into one. 

In addition to the above, allowance is made for wave action, 
the height of which is obtained by the following formula: 

^ = 1.5V?+(2.5- y¥) ( 4 ) 

In this expression F is the “fetch”, or longest line of exposure of 
the water surface to wind action in miles. Thus if F = 4 miles, the 
extra height required over and above maximum flood level will be 
(1.5X2)+ (2.5 —1.4) =4.1 feet. If F=10 miles, h w will work out to 
5| feet. The apex of the triangle of water pressure must be placed 
at this higher level; the crest, however, is frequently raised still 
higher, so as to prevent the possibility of water washing over it. 

14. Example. The working out of an actual example under 
assumed conditions will now be given by both graphical and analyti¬ 
cal methods. Fig. 8 represents a profile 50 feet in height with crest 
level corresponding with the apex of the triangle of water pressure. 


DAMS AND WEIRS 


15 


The assumed value of p is 2\. The outline is nearly pentagonal, 

H 2 

the crest width is made .155 and the base width is the full 

Vp ° 

X50 = 33.3 feet, the crest width is thus 5 feet. The back slope is 
carried down vertically to the point e, a distance of 8 feet, and from 
here on, it is given a batter of 1 in 50. The outset at the heel beyond 
the axis of the dam, which is a vertical line drawn through the rear 



edge of the crest is therefore .84 foot * The toe is set in the same 
extent that the heel is set out. The face line of the body is formed 
by a line joining the toe with the apex of the water-pressure triangle. 
If the face line were drawn parallel to the hypothenuse of the ele¬ 
mentary triangle, i.e., to a slope of 1 : Vp, it would cut off too much 
material, the area of the wall being then but very little in excess of 
that of the elementary triangle, which, of course, is a minimum 
quantity. As will be seen later, the analysis of the section will show 
that the adopted base width could have been reduced below what 
































16 


DAMS AND WEIRS 


has been provided, to an extent somewhat in excess of that given in 
formula (3). 

15. Graphical Method. The graphical procedure of drawing 
the resultant lines W (R.E.) and R (R.F.) to their intersection of the 
base presents a few differences, from that described in section 7, 
page 6, with regard to Fig. 3. Here the profile is necessarily divided 
into two parts, the rectangular crest and the trapezoidal body. As 
the three areas (1), (2), and Pi, are not of equal height, the item H 
cannot be eliminated as a common factor, consequently the forces 
will have to be represented as in Fig. 6 by their actual superficial 
areas, not by the half width of these areas as was previously the case. 
In Fig. 8a the vertical load line consists of the areas 1 and 2 totaling 
844 square feet, which form IF. The water pressure Pi is the area 

of the inclined triangle whose base is —. This is best set out graph- 

P 


ically in the force polygon by the horizontal line P, made equal to 


the horizontal water pressure, which is 


H 2 2500X2 


= 555 square 


2p 9 

feet. The water-pressure area strictly consists of two parts corre¬ 
sponding in depth to (1) and (2) as the upper part is vertical, not 
inclined, but the difference is so slight as to be inappreciable, and 
so the area of water pressure is considered as it would be if the back of 
the wall were in one inclined plane. In Fig. 8 the line Pi normal to 
the back of the wall is drawn from the point of origin 0 and it is cut 
off by a vertical through the extremity of the horizontal line P. 
This intercepted length 0 0 1 is clearly the representative value of 
the resultant water pressure, and the line joining this point with the 
base of the load line W is P, the resultant of IF and of P x . If a 
horizontal line AB be drawn from the lower end of the load line IF 
it will cut off an intercept ( N ) from a vertical drawn through the 
termination of Pi. This line AB = P, and N is the vertical com¬ 
ponent of P, the latter being the resultant of IF and Pi as well as 
of N and P. When the back is vertical, N and IF are naturally 
identical in value, their difference being the weight of water over- 
lying the inclined rear slope. 

The further procedure consists in drawing the reciprocals of 
the three forces Pi, IF, and P on the profile. The first step consists 
in finding the centers of gravity of the vertical forces 1 and 2 in which 




DAMS AND WEIRS 


17 


the hexagonal profile is divided. That of (1) lies clearly in the 
middle of the rectangle whose base is de. The lower division (2) is 
a trapezoid. r lhe center of gravity of a trapezoid is best found by 
the following extremely simple graphical process. From d draw 
dli horizontally equal to the base of the trapezoid fg and from g, 
gj is set oft* equal to de; join hj, then its intersection with the middle 
line of the trapezoid gives the exact position of its center of gravity. 
Thus a few lines effect graphically what would involve considerable 
calculation by analytical methods, as will be shown later. 

The next step is to find the combined c. g. of the two parallel 
and vertical forces 1 and 2. To effect this for any number of parallel 
or non-parallel forces, two diagrams are required, first, a so-termed 
force and ray polygon and, second, its reciprocal, the force and 
chord, or funicular polygon. The load line in Fig. 8a can be utilized 
in the former of these, figures. First, a point of origin or nucleus of 
rays must be taken. Its position can be anywhere relative to. the 
load line, a central position on either side being the best. The 
point 0\ y which is the real origin of the force polygon at the extremity 
of Pi can be adopted as nucleus and often is so utilized, in which 
case the force line Pi and R can be used as rays, only one additional 
ray being required. For the sake of illustration, both positions for 
nucleus have been adopted, thus forming two force and ray poly¬ 
gons, both based on the same load line, and two funicular polygons, 
the resultants of which are identical. The force and ray polygon is 
formed by connecting all the points on the load line with the nucleus 
as shown by the dotted line a, b, and c, and a', b', and c'. Among the 
former, a and c are the force lines Pi and P, the third, b, joins the 
termination of force (1) on the load line with the nucleus. The§e 
lines a, b, c , are the rays of the polygon. Having formed the force 
and ray diagram, in order to construct the reciprocal funicular 
polygon 8b the force lines (1) and (2) on the profile Fig. 8 are con¬ 
tinued down below the figure. Then a line marked (a) is drawn 
anywhere right through (1) parallel to the ray a, from its intersec¬ 
tion with the force (1), the chord (b) is drawn parallel to the ray ( b ) 
in Fig. 8b meeting (2); through this latter intersection the third 
chord (c) is drawn backward parallel to its reciprocal the ray c. 
This latter is the closing line and its intersection with the initial 
line (a), gives the position of the c.g. of the two forces. 


18 


DAMS AND WEIRS 


A vertical line through this center of pressure, which represents 
W, i.e., W 1 +W 2 , is continued on to the profile until it intersects 
the inclined force Pi drawn through the center of gravity of the 
water pressure area. This intersection is the starting point of P, 
drawn parallel to its reciprocal on the force polygon 8a. This 
resultant intersects the base at a point within the middle third. 
R is the resultant “Reservoir Full”, while W, the resultant of the 
vertical forces in the masonry wall, is the resultant “Reservoir 
Empty”. The intersection of the latter is almost exactly at the inner 
edge of the middle third—thus the condition of the middle third is 
fulfilled. The question of induced pressure and its distribution on 
the base will be considered later. 

The incidence of N, the vertical component “Reservoir Full”, 
on the base is naturally not identical with that of IF, the resultant 
“Reservoir Empty”, unless the back of the wall is vertical. The 
line R is the resultant of both Pi and IF, and of P and N. If 
it be required to fix the position of N on the profile, a horizontal 
line should be drawn through the intersection of Pi with the back 
of the wall. This will represent the horizontal component of the 
water pressure Pi, and it will intersect R, produced upward. Then 
a line drawn vertically through this latter point will represent N, 
the vertical component (Reservoir Full). The position of N is 
necessarily outside of IF, consequently if N is made to fall at the inner 
edge of the middle third of the base, IF must fall within the middle 
third. This fact will later be made use of when the design of the 
lower part of a “high” dam comes under consideration. 

16. Analytical Method. The analytical method of ascertain¬ 
ing the positions of the incidences of IF and of R on the base, which 
has just been graphically performed, will now be explained. 

The first step is to find the positions of the centers of gravity 
of the rectangle and trapezoid of which the profile is composed, 
relative to some vertical plane, and then to equate the sum of the 
moments of those two forces about any fixed point on the base, 
with the moment of their sum. 

The most convenient point in most cases is the heel of the base; 
this projects a distance ( y ) beyond the axis of the dam, which axis 
is a vertical line passing through the inner edge of the rectangular 
crest. 


DAMS AND WEIRS 


19 


As the areas of the divisions, whether of the masonry wall or 
of the water-pressure triangle, are generally trapezoids, the follow¬ 
ing enumeration of various formulas, 
whereby the position of the c.g. of a 
trapezoid may be found either with regard 
to a horizontal or to a vertical plane, 
will be found of practical utility. In Fig. 

9, if the depth of the figure between the 
parallel sides be termed //, and that of 
the truncated portion of the triangle of 
which the trapezoid is a portion be 
termed d, and h be the vertical height 
of the c.g. above the base, then 



Fig. 9. Diagram Showing Centers 
of Gravity of Water Pressure 
Trapezoids 


H H+Sd 
3 X H+2d 



Thus, in Fig. 9, H = 13 and d = 6 feet, then 



/ 13-f 18 \ 
\13-fl2/ 


= 5.37 feet 


If the base of the triangle and trapezoid with it be increased or 
decreased in length, the value of h will not be thereby affected, 
as it is dependent only on H and d, which values are not altered. 
If, however, the base of the triangle be inclined, as shown by the 




Fig. 10. Diagram Illustrating Height of c. g. Trapezoid above Base 


dotted lines in Fig. 9, the center of gravity of the trapezoid will 
be higher than before, but a line drawn parallel to the inclined 
base through g, the c. g. will always intersect the upright side of 
the trapezoid at the same point, viz, one which is h feet distant 
vertically above the horizontal base. 























20 


DAMS AND WEIRS 


The value of h can also be obtained in terms of a and b, the two 
parallel sides of the trapezoid, and is 

'b-\-2a 


*-! 


/6+2a\ 

\a+b) 


( 6 ) 


For example, in Fig. 10, II = 12, a = 10, and 6 = 16, then 

'16+20' 


h 


_ 12 /16+20\ 
: 3 VlO+16/ 


= 5.54 feet 


. 10 + 1 + 

If the horizontal distance of the c. g. of a trapezoid from a ver¬ 
tical plane is required, as, for example, that of the trapezoid in Fig. 
8, the following is explanatory of the working. As shown in Fig. 
11, this area can be considered as divided into two triangles, the 
weight of each of which is equivalent to that of three equal weights 
placed at its angles; each weight can thus be represented by one- 

third of the area of the triangle in question, or by and respec- 

6 6 

tively, II being the vertical depth of the trapezoid. Let y be the 

projection of the lower corner 
A beyond that of the upper 
one B. Then by equating 
the sum of the moments of 
the corner weights about the 
point A with the moment of 
their sum, the distance (.t) of 
the c.g. of the whole trape¬ 
zoid from A will be obtained 
as follows: 

OlT’) = Y [ & (a+6)+a(a+i/)+2/(a+&)l 

*=t[<w(3)] 

where y = 0, the formula becomes 



Fig. 11. Method of Finding Distance of Center 
of Gravity of a Trapezoid from Heel 


( 7 ) 


X 


.l( b+ J±.\ 

3 V a+b) 


(7a) 


For example, in Fig. 8, a or de = 5 feet, 6 = 33.3, and y = M, whence 

29.2' 


”T( 34 ' 14+ H)- 1I - 63 '«' t 





















DAMS AND WEIRS 


21 


The similar properties of a triangle with a horizontal base, as 
in Fig. 12, may well be given here and are obtained in the same way 
by taking moments about A, thus 


bh bh n . 
Y Xx== ~b ^+y) 


x = 


b+y 


In Fig. 12, 5 = 14 feet, y = 8 feet, and h = 10 feet, then 

W-= TWeet 


X 



Reverting to Fig. 8, the position of the incidence of IF on the 
base is obtained by taking moments about the heel g of the base as 
follows: Here IF is the area of the whole profile, equal, as we have 
seen, to 844 sq. ft. The area of the upper component (1) is 40 sq. 
ft. and of (2) 804. 

The lever arm of IF is by hypothesis x, 
that of (1) is 2.5 + .84 = 3.34 feet, that of (2) 
by formula (7) has already been shown to be 
11.63 feet. Hence, as the moment of the whole 
is equal to the sum of the moments of the bh 
parts, the equation will become 



844.r = 40 X 3.34+804 X11.63 = 9484.1 


Fig. 12. Method of 
Finding Center of 
Gravity of a Triangular 
Profile 


x — 11.23 feet 


This fixes the position of the incidence of W relative to the heel. 

b 33 3 

The position of the inner third point is or ~tt~ ^ rom ^ ie ^ ee ^‘ 

The incidence of IF is therefore 11.23-11.10 = .13 foot within the 

middle third, which complies with the stipulated proviso. 

The next step is to find the position of R relative to the heel of 

the base. As in graphical methods, only horizontal and vertical 

forces are considered; the water-pressure area is split into two parts, 

H 2 

one, P the horizontal component, the value of which is —, or 555 

feet, and w 3 the reduced area of water overlying the rear projection 
of the back. The latter area is a trapezoid of which the upper side 














22 


DAMS AND WEIRS 


(a) is 8 feet long and b, the lower side, 50 feet, the depth being .84 
foot, hence the distance of its c. g. inside the heel of the base will be 

by formula (6), = .32 foot. Its actual area is 

.84 = 24.4 feet; this has to be divided by p or 2J to reduce it to a 
masonry base. The reduced area will then be 10.8 square feet, 
nearly. The distance of the incidence of W from the heel of the 
base has already been determined to be 11.23 ft. and that of Wz 
being .32 ft., the distance of the c. g. of the latter from W will be 
11.23 —.32 = 10.9, nearly. If the distance between the incidences of 
W and R be termed x, the equation of moments about the incidence 
of R, will stand thus: 


or 


i.e. 


P— = W x-\-Wz(x-{- 10.9) 

O 


555 X— = 844*+10.8*+'l 17.83 

o 


9132.2 

854.8 


= 10.7 ft., nearly 


R is therefore 10.7+11.23 = 21.93 ft. distant from the heel. The 
§ point being 22.2 ft. from the same point, R falls .3 ft. (nearly) 
within the middle third. This shows that a small reduction in the 
area of the profile could be effected. 

17. Vertical Component. If the position of N, the vertical 
component of R and Pi, is required, as is sometimes the case, 
it is obtained by the equation iVX£=(IFXll.23) + (w 3 X.32), 
x being the distance from the heel of the base. Or in figures, 

854.8.r = (844 X 11.23) + (10.8 X .32) 


£ = 11.1 feet 

The incidence of N is, therefore, in this case, exactly on the limit of 
the middle third. This of course does not affect the condition of 
middle third, which refers to the resultant W (R.E.) not to the com¬ 
ponent N (R.F.) but, as will be seen later, when the lower part of a 
high dam comes to be designed, one condition commonly imposed 
is, that the vertical component N must fall at the inner edge of the 
middle third, in which case W will necessarily fall inside thereof. 






DAMS AND WEIRS 


23 


It may here be noted that the space between the location of A and 

P H 

R, which will be designated (/), is 7 ^ because if moments are taken 

o iV 

PH PH 

about the incidence of R, then Nf= -; therefore /=-—-. The 

actual value of lb in tons of 2000 pounds will be the superficial area, 
or 844 square feet multiplied by the eliminated unit weight, i.e., by 
344 ><9 1 

wp, viz, — - = 59.3 tons, as w = — ton. That of the inclined 

oZX4 6Z 


force R , is obtained from the triangle of forces PAR in which P, 

being the hypothenuse =ViV 2 +P 2 . Here A = 855 square feet, 
equivalent to 60 tons, nearly, and P = 555 feet, equivalent to 39 

tons, whence R = V60 2 +39 2 = 71.5 tons. 

18. Pressure Distribution. In the design of the section of a 
dam, pier, or retaining wall, the distribution of pressure on a plane 
in the section and the relations existing between maximum unit 
stress, symbolized by (5), and mean or average unit stress (si) will 
now be considered. The mean unit stress on any plane is that 
which acts at its center point and is in amount the resultant stress 
acting on the plane (the incidence of which may be at any point) 
divided by the width of the lamina acted on. Thus in Figs. 3 or 8 
take the resultant Ilk This acts on the horizontal base and its 

W 

mean unit stress Si will be —. In the same way, with regard to A, the 

vertical component of R the mean unit stress produced by it on the 

A 

horizontal base will be —. The maximum unit stress occurs at 

0 


that extremity of the base nearest to the force in question which is 
R. Thus the maximum unit stress due to W is at the heel while 
that due to a combination of P and A acting at the incidence of R 
is at the toe of the base b. It is evident that the nearer the inci¬ 
dence of the center of pressure is to the center point the less is the 
maximum stress developed at the outer edge of the section, until 
the center of pressure is actually situated at the center point itself. 
The maximum pressure at the outer part of the section then equals 
the average and is thus at a minimum value. The relation between 
maximum and mean unit stress or reaction is expressed in the fol- 








24 


DAMS AND WEIRS 


lowing formula in which it is assumed that any tension at the heel 
can be cared for by the adhesion of the cementing material or of 
reinforcement anchored down: 

« = *i(l+y) (9) 

or, letting m equal the expression in brackets, 

s = ms I (9a) 


In formula (9a), q is the distance between the center point of the 
base and the center of pressure or incidence of whatever resultant 

pressure is under considera¬ 
tion, and Si is the mean stress, 
or the resultant pressure di¬ 
vided by the base. 

In Fig. 8 as explained in 
section 16, the incidence of R, 
i.e., the center of pressure 
(IFF.), falls .3 ft. within the 
middle third of the base, con¬ 
sequently the value of q will be 



6 


.3 = 


33.3 

6 


.3=5.25 ft., and 

in formula (9a) m = 1 + ~ 

b 


= 1 


31.5 


33.3 

mum reaction 


= 1.95. The inaxi- 
at 


always designated by s = 


the toe 
mN 


Fig. 13. Diagram Showing Pressure Distribution 
on a Dam with Reservoir Empty and 
Reservoir Full 


1.95X60 

33.3 


= 3.51 tons per sq. 


ft. For the reaction (R.F.) at 

the heel, m = 1 - .95 = .05, and s 2 = = .09 tons. The distribu- 

<)>) .5 

tion of pressure due to the vertical component of R is shown hatched 
in Fig. 8 as well as in Fig. 13. 

From formula (9) the facts already stated are patent. When 
the incidence of the resultant force is at the center of the base, 


































DAMS AND WEIRS 


25 


q 0, consequently m — 1 and s — s\, that is, the maximum is equal 

to the mean; when at one of the third points, q — m — 2, and s = 2s \; 

6 

TT/ 

when at the toe, m = 4, and s = 4s u or 4—. 

b 

If the material in the dam is incapable of caring for tensile 
strain, the maximum vertical compression, or s, obtained by formula 
(9) will not apply. Formula (24), section 86, should be used when¬ 
ever R falls outside the middle third. 

In designing sections it is often necessary to maneuver the 
incidence of the resultant stress to a point as close as possible to the 
center of the base in order to reduce the maximum stress to the least 
possible value, which is that of the mean stress. The condition of the 
middle third* insures that the maximum stress cannot exceed twice 
the mean, and may be less, and besides insures the absence of tensile 
stress at the base. 

19. Graphical Method for Distribution of Pressure. The 
graphical method of ascertaining the distribution of pressure on the 
base of a masonry wall, which has already been dealt with analyti- 
cally, is exhibited in Fig. 13, which is a reproduction of the base of 
Fig. 8. The procedure is as follows: Two semicircles are struck 
on the base line, having their centers at the third division points 

and their radii equal to From the point marked e, that of the 

O 

incidence of R, the line eg is drawn to g, the point of intersection of 
the two semicircles. Again from g a line gn is set off at right 
angles to eg cutting the base or its continuation at a point n. This 
point is termed the antipole of e, or the neutral point at which pres¬ 
sure is nil in either sense—compressive or tensile. Below and 
clear of the profile a projection of the base is now made, and from 
g a perpendicular is let fall, cutting the new base in g' while, if the 
line be continued upward, it will intersect the base at K. This 
latter point will, by the construction, be the center point of the base. 
The line Kg is continued through g' to h', g'h' being made equal to 
the mean unit pressure, =1.8 tons. A perpendicular is let fall 
from n cutting the new base line at n\\ the points n\ and h' are then 
joined and the line continued until it intersects another perpendicu¬ 
lar let fall from the toe of the base. A third perpendicular is drawn 


26 


DAMS AND WEIRS 


from the heel of the base, cutting off a corner of the triangle. The 
hatched trapezoid enclosed between the last two lines represents the 
distribution of pressure on the base. The maximum stress will 
scale close upon 3.51 and the minimum .09 tons. If W be considered, 
W 59 3 

s = — = —— = 1.78 tons, the maximum stress at the heel will be 
b 33.3 

3.52 and the minimum .04, at the toe. 

20 . Examples to Illustrate Pressure Distribution. In Fig. 14 
is illustrated the distribution of pressure on the base, due to the 

incidence of R, first, at the toe of 
the base, second, at the two-third 
point, third, at the center, and 
fourth, at an intermediate position. 
In the first case ( Ri ), it will be 
seen that the neutral point n\ falls 
at the first third point. Thus two- 
thirds of the base is in compression 
and one-third in tension, the maxi¬ 
mum in either case being propor¬ 
tional to the relative distance of the 
neutral point from the toe and heel 
of the base, the compression at the 
toe being four times, while the ten¬ 
sion at the heel is twice the mean 
stress. In the second case (R 2 ) 
intersects at the two-third point, 
and the consequent position of n is 
exactly at the heel. The whole base 
is thus in compression, and the max¬ 
imum is double the mean. In the 
third case (7? 3 ), the line gn is 
drawn at right angles to fg. The 
latter is vertical and gn will conse¬ 
quently be horizontal. The distance to n is thus infinite and the 
area of pressure becomes a rectangle with a uniform unit stress s. 
In the fourth case (R 4 ), the neutral point lies well outside the profile, 
consequently the whole is in compression, the condition approximating 
to that of I? 3 . 



Fig. 14. Pressure Distribution on 
Base of Dam under Various 
Conditions 




















DAMS AND WEIRS 


27 


21. Maximum Pressure Limit. The maximum pressure 
increases with the depth of the profile until a level is reached where 
the limit stress or highest admissible stress is arrived at. Down to 
this level the design of the section of a dam, as already shown, con¬ 
sists simply in a slight modification of the pentagonal profile with 
a vertical back, the base width varying between that of the ele¬ 
mentary profile or or its reduced value given in formula ( 3 ). 

yip 


Beyond this limiting depth, which is the base of the so-termed 
“low” dam, the pentagonal profile will have to be departed from 
and the base widened out on both sides. 

22. Formulas for Maximum Stress. The maximum unit stress 
in the interior of a dam is not identical with ( s ), the maximum vertical 
unit reaction at the base, but is a function of Si. In Fig. 8 , a repre¬ 
sentative triangle of forces is shown composed of N the vertical 
force (R.F.), P the horizontal water pressure, and R the resultant 

of N and P; therefore R = V jV 2 +P 2 = also N sec 0 . If the back 

were vertical, N and W would coincide and then P = VlF 2 -fP 2 . 
Various views have been current regarding the maximum internal 
stress in a dam. The hitherto most prevalent theory is based on 
the assumption, see Fig. 8 , that the maximum unit stress 


m R V^ 2 +P 2 

c = 7 = m - - - 

b b 


mN 


seed 


(10a) 


Another theory which still finds acceptance in Europe and in the 
East assumes that the maximum stress is developed on a plane 
normal to the direction of the resultant forces as illustrated by the 
stress lines on the base of Fig. 8 . According to this, the mean 

R R 

stress due to R would not be — but —, and the maximum stress will 

b bi 

be PlIL' But = ^ sec — an( i R= N sec 6, consequently the maxi- 

61 b 1 b 

mum unit stress would be 

c = ^7^-sec 2 0 (10b) 

b 


Recent experiments on models have resulted in the formula for 
maximum internal unit stress being recast on an entirely different 
principle from the preceding. The forces in action are the maxi- 












28 


DAMS AND WEIRS 


mum vertical unit force or reaction s combined with a horizontal 

P 

shearing unit stress s s =—. The shearing force is the horizontal 

b 


water pressure, or 


H 2 w 


symbolized by P, which is assumed to be 

*/ t j * 


equally resisted by each unit in the base of the dam; the unit shear- 

P 

ing stress will thus be —. These forces being at right angles to 

b 

each other, the status is that of a bar or column subject to compres¬ 
sion in the direction of its length and also to a shear normal to its 
length. The combination of shear with compression produces an 
increased compressive stress, and also a tension in the material. 
The formula recently adopted for maximum unit compression is 
as follows: 

( 10 ) 


c = \s-\- Vi.s 2 +S s 2 


In this s = msit = 


m N 


P 

As before s s =—, substituting we have 

b 


m N 


2b ’’M 46 2 ' b 2 2b 

When m = 2, as is the case when the incidence of R is exactly at the 
outer boundary of the middle third 

a+Vf+f n 


(vi N ) 2 P 2 vi A r + V(mA) 2 +4P 2 


(10J 


c — 


(1+ sec 6) 


( 10 2 ) 


b b 

23. Application of All Three Formulas to Elementary Profile. 

In the case of elementary triangular profile which has a vertical 


back, N = W and sec 6 = 


y>+i 


V/ 


(section 7, page 5) and m = 2; then 


formula (10 2 ) becomes 




P + 1 


) 


Now 


1F_ gj Vp_g W p 
b 2Vp Xp X H 2 


c = 


Hwp 


( 


i+ %iy) 


(ii) 























DAMS AND WEIRS 


29 


Example. 

Let II in elementary triangle = 150 feet, p = 2.4, wp = — ton. 

When, according to (11), c-- 1 f' 3 (1 + 1.187) = 12.3 tons per 
square foot. 

Taking up formula (10a) 

mW sec 6 2 W sec 6 


c = 


, W Hwp 

as above — =-- 

b 2 


= Hwpyj^^ = #wVpVp+l 


(Ha) 


Example with conditions as before 


c = 


150X1XI. 55V3.4 


32 


= 7.26 X 1.84 = 13.3 tons 


'With formula (10b), c— ^ S * 6C ^ , or in terms of II, 


c= IIwp 


(?)- 


Hw(p+ 1 ) 


(Hb) 


Therefore, with values as above, 

150X1X3.4 


c = 


OO 

oZ 


= 15.9 tons 


From the above it is evident that formulas (10b) and (lib) give 
a very high value to c. Tested by this formula, high American 
dams appear to have maximum compressive unit stresses equal to 
20 tons per square foot, whereas the actual value according to 
formula (10) is more like 14 tons. However, the stresses in the 
Assuan dam, the Periyar, and other Indian dams, as also French 
dams have been worked out from formula (10b) which is still in use. 

24. Limiting Height by Three Formulas. The limiting height 
(III) of the elementary triangular profile forms a close guide to 
that obtaining in any trapezoidal section, consequently a formula will 
be given for each of the three cases in connection with formulas (10), 
(10a), and (10b). Referring to case (10), we have from formula (11) 


















30 


DAMS AND WEIRS 


Whence Hi, the limiting height = 


2c 




wp 


Example. 

With c = 16 tons and p = 2.4, Hi, the limit height of the ele- 

t 2X16X33 1024 inK t + 

mentary profile will be - = 195 feet. 

Referring to case (I0a), we have from formula (11a) 

c= Hw VpVp + l 


H,= 


Example. 

With data as above H = 


w VpVp+i 
16X32 


= 180 feet, nearly. 


1.55X1.84 

Referring to case (10b), we have from formula (lib) 

c= Hiv(p-\-1 ) 


Example. 



c 

w(p+ 1) 


With same data Hi = 


16X32 

3.4 


512 

3.4 


= 150 feet. 


Thus the new formula (10) gives much the same results as that 
formerly in general use in the United States (10a), while in the 
more conservative formula (10b) the difference is marked. 

25. Internal Shear and Tension. We have seen that the 
combination of compressive and shearing stresses in a dam (R.F.) 
produces an increased unit compression. It further develops an 
increase in the shearing stress and also a tensile stress. The three 
formulas are given below. 

Compression as before 


1 

c = —5 + 


s* „ m 

— +s s 2 or — 
4 


A r +A) 2 +4P 2 / jq\ 

2b ^ ^ 


Tension 


1 I s 2 

= y s At 


-f«? s 2 or 


m N — V (m N) 2 +4P 2 


26 


( 12 ) 





















DAMS AND WEIRS 


31 


Shear 


Sh = 



or 


(miV) 2 +4P 2 


2 b 



The tensile and shearing stresses are not of sufficient moment to require 
any special provision in the case of a gravity dam. The tension is 
greatest at the heel, diminishing toward the toe. This fact suggests 
that a projection of the heel backward would be of advantage. The 
direction (a) of c to the vertical is not that of R but is as follows: 


Tan 2a 
P 


2 s 8 _ 2P mN _ 2P 
s b b m N 


when m = 2, tan 2a = —. In Fig. 8, P = 555 and A 7 = 855. /.tan 2a = 
555 

^^ = .649 whence 2a = 33° 00 r and (a) = 16° 30'. The inclination 


of R to th6 vertical, or 8, is 33° 50 r , i.e., twice as large as that 
of c. The direction of t is at right angles to that of c, while that of 
Sh the shear, lies at 45° from the directions of either c or t. 

26. Security against Failure by Sliding or Shear. Security 
against failure by sliding depends on the inclination of W to P, i.e., 
on the angle 8 between W and R. Thus tan 8 should be less than the 
angle of friction of masonry on masonry, or less than .7. This is the 
same as stating that the relation of W and P must be such that 8 
shall not be greater than 35°, or that the complement of 8 be not less 
than 55°. The adoption of the middle third proviso generally insures 
this. With regard to sliding on the base, this can be further pro¬ 
vided against by indentations in the base line or constructing it 
inclined upward from heel to toe. 

27. Influence Lines. It is sometimes desirable for the purpose 
of demonstrating the correctness of a profile for tentative design, 
to trace the line of pressures corresponding to the two conditions 
of reservoir full and empty, through the profile of a dam. This is 
far better effected by the use of graphic statics. 

There are two different systems of graphic construction that 
give identical results, which will now be explained and illustrated. 

The first method, which is most commonly adopted, is exhibited 
in Fig. 15, which is the profile of a 100-foot high dam with specific 
gravity 2\. It thus lies within the limiting depth, which for the 
elementary profile would be 190 feet. 










32 


DAMS AND WEIRS 


The profile is pentagonal, with a vertical back, and has the full 
base width of the elementary profile, viz, which in this case is 

Vp 


100 = 66.7 feet. The crest k is Vi/=10 feet wide. The water- 

O 

pressure triangle has a base of —. The profile, as well as the water- 

P 

pressure triangle, is divided into five equal laminas, numbered 1 



Fig. 15. Graphical Construction for Tracing Lines of Pressure on Dam of Pentagonal Profile 


to 5 in one case and 1' to 5' in the other. The depth of each lamina, 

which is ~ is, therefore, a common factor and can be eliminated as 
5 

well as the item of unit weight, viz, wp. The half widths of all 
these laminas will then correctly represent their areas and also their 
weights, reduced to one denomination, that of the masonry. In 
Fig. 15a a force polygon is formed. In the vertical load line the 
several half widths of the laminas 1 to 5 are first set off, and at 




















































DAMS AND WEIRS 


33 


right angles to it the force line of water pressure is similarly set 
out with the half widths of the areas 1' to 5 r . Then the resultant 
lines ol the combination of 1 with 1', 1,2, with 1' 2' and so on marked 
Ri to /?5 are drawn. This completes the force polygon. The next 
step is to find the combinations of the vertical forces on the profile, 
viz, that of 1 and 2, 1, 2, and 3, etc. This, as usual, is effected by 
constructing a force and ray polygon, utilizing the load line in 
Fig. 15a for the purpose. Then the centers of gravity of the several 
individual areas 1 to 5 are found by the graphical process described 
in section 15, and verticals drawn through these points are projected 
below the profile. On these parallel force lines 1 to 5, the funicular 
polygon Fig. 15b is constructed, its chords being parallel to their 
reciprocal rays in Fig. 15a. The intersection of the closing lines 
of the funicular gives the position of the centroid of the five forces 
engaged. By producing each chord or intercept backward until it 
intersects the initial line, a series of fresh points are obtained which 
denote the centers of gravity of the combinations of 1 and 2; 1,2, and 
3, and so on. Verticals through these are next drawn up on the profile 
so as to intersect the several bases of the corresponding combinations, 
thus 1, 2, and 3 will intersect the base of lamina 3; and 1, 2, 3, and 
4 will intersect that of lamina 4; and so on. These intersections 
are so many points on the line of pressure (R.E.). The next step 
is to draw the horizontal forces, i.e., their combinations on to the 
profile. The process of finding the centers of gravity of these areas 
is rendered easy by the fact that the combinations are all triangles, 
not trapezoids, consequently the center of gravity of each is at J 
its height from the base. Thus the center of gravity of the com¬ 
bination l / +2 / +3 / is at J the height measured from the base of 
3' to the apex, in the same way for any other combination, that of 
1 ', 2', 3 r , 4', 5being at J the total height of the profile. The back 
being vertical, the direction of all the combined forces will be hori¬ 
zontal, and the lines are drawn through, as shown in the figure, 
to intersect the corresponding combinations of vertical forces. 
Thus V intersects 1 , 1' 2' intersects 1 2 , and so on. From these 
several intersections the resultant lines 7?i, i?2, to R$ are now drawn, 
down to the base of the combination to which they belong, these 
last intersections giving the incidence of Ri, R 2 , etc., and are so 
many points on the line of pressures (R.F.). The process is simple 


34 


DAMS AND WEIRS 


and takes as long to describe as to perform, and it has this advan¬ 
tage, that each combination of forces is independent of the 
rest, and consequently errors are not perpetuated. This system 
can also be used where the back of the profile has one or 
several inclinations to the vertical, explanation of which will be 
given later. 

28. Actual Pressures in Figures. In the whole process above 
described, it is noticeable that not a single figure or arithmetical 
calculation is required. If the actual maximum unit stress due to 
R or to IF is required to be known, the following is the procedure. 
In Fig. 15a, N scales 174, to reduce this to tons it has to be mul¬ 
tiplied by all the eliminated factors, which are -^ = 20 and wp = 


9X1 Ar 174X20X9 

4X32’ that 1S ’ N = 4x32"" = 244 t0nS ' 

Assuming the incidence of R exactly at the third division, 

the value of q is — and that of m is 2; P also scales 112, its 

b 

value is therefore = 157 tons. Applying formula (10 2 ), 

4 X oJ 


„ , ViV 2 +P 2 244+V244 2 +157 2 534 _ ^ 

c = A H---=--=7^777 = 8 tons per sq. it., 


66.7 


66.7 


roughly. As 8 tons is obviously well below the limiting stress, for 
which a value of 16 tons would be more appropriate, this estimation 
is practically unnecessary but is given here as an example. 

29. Analytical Method. The analytical method of calculation 
will now be worked out for the base of the profile only. First the 
position of IF, the resultant vertical forces (R.E.) relative to the 
heel of the base will be calculated and next that of R. The back of 
the profile being in one line and vertical the whole area can be con¬ 
veniently divided into two right-angled triangles, if the thick¬ 
ening of the curvature at the neck be ignored. As the fore slope 

has an inclination of 1 :Vp the vertical side of the upper triangle 


(1) is /cVp in length; its area will then be 


Wp 100X1.5 


= 75 


sq. feet. The distance of its c. g. from the heel of the base, which 

20 

in this case corresponds with the axis of the dam, is — feet = 6§ feet. 

O 












DAMS AND WEIRS 


35 


er 


75 V 20 

r lhe moment will then be —-— = 500. With regard to the low 

O 

H 2 2 

triangle, its area is -—-= = 5000 X- = 3333.3 sq. feet. The length of 

2 Vp 3 

its lever arm is one-third of its base, or 22.2 feet. The moment 
about the axis will then be 3333.3X22.2 = 74,000. The moment of 
the whole is equal to the sum of the moments of the parts. The 
area of the whole is 75+3333 = 3408. Let x be the required distance 
of the incidence of W from the heel, then 

zX 3408 = 74,500 
74,500 


x — 


3408 


= 21.9 feet 


The inner edge of the middle third is — or 22.2 feet distant from the 

O 

heel; the exact incidence of W is, therefore, .3 foot outside the middle 

third, a practically negligible amount. With regard to the position 

of R the distance (/) between the incidence of R and that of W is 

PH • +l . p+l , 100X44.4 ooon . , 

—j-p-; in this R the water-pressure area=-— --=2220. . . j = 

- = 21.7 feet. The total distance of R from the heel will 

3 X 3408 

then be 21.7+21.9 = 43.6 feet; the outer edge of the middle third 
is 44.4 feet distant from the heel, consequently the incidence of R 

j rr 7 

is 44.4 — 43.6 = .8 foot within the middle third, then q = ——.8=—+ 

6 6 

— .8 = 10.3 ft., and ra = ^1+—^ = 1 + .93 = 1.93. At this stage it 

will be convenient to convert the areas into tons by multiplying 

2.25 


them by pw, or 


32 


Then N and W become 239.6, and P becomes 


156.3 tons. Formula (10) will also be used on account of the high 
figures; then 


Here s = 


mW 239.6X1.93 


s £ 


+s« 


66.7 


= 6.93 tons, and s s = 


P_ 156.3 
b 66.7 


= 2.34 


,, , 6.93 

tons, therefore, c = ——h \ 


6.93 


f (2.34) 2 = 3.46+Vl7.48 = 7.64 tons. 





















36 


DAMS AND WEIRS 


, , , , 6</ 239.6X0.07 

For 5 2 , or the compression at the heel, m— 1 — — = .U /. 5 2 — 0 ^ 7 ” 

= .251 ton. The area of base pressure is accordingly drawn on Fig. 15. 

If IF (R.E.) be considered, <7 =—+.3 = 11.42, and m = 1 + 

6 

6X11.42 . , mW 239.6X2.03 „ OA , ^ 

———— =2.03; therefore, 5 = -=— =-777^-=7.30 tons. The 

66.7 b 66 ./ 

base pressure is therefore greater with (R.E.) than with (R.F.); 



Fig. 16. Diagram Showing Haessler’s Method for Locating Lines of Pressure on a Dam 


there is also a slight tension at the toe of .11 ton, a negligible 
quantity. This pressure area is shown on Fig. 16. 

30. Haessler’s Method. A second method of drawing the 
line of pressures which is termed “Haessler’s” is exhibited in Fig. 
16, the same profile being used as in the last example. In this 
system, which is very suitable for a curved back, or one composed 
of several inclined surfaces, the forces are not treated as independent 
entities as before, but the process of combination is continuous 
from the beginning. They can readily be followed on the force 
polygon, Fig. 16a and are F with 1 producing Ri; Ri with 2', i.e., 
F, 1, 2', the last resultant being the dotted reverse line. This last 
is then combined with 2 producing R 2 , and so on. 


















































DAMS AND WEIRS 


37 


The reciprocals on the profile are drawn as follows: First the 
c.g.’s of all the laminas 1, 2, 3, etc., 1', 2', 3', etc., are obtained by 
graphical process. Next the water-pressure lines, which in this case 
are horizontal, are drawn through the profile. Force line (F) inter¬ 
sects the vertical (1), whence R\ is drawn parallel to its reciprocal 
in Fig. 16a through the base of lamina (1) , until it reaches the hori¬ 
zontal force line (2'). Its intersection with the base of (1) is a point 
in the line of pressure (R.F.). Again from the intersection of R\ 
with (2'), a line is drawn backward parallel to its dotted reciprocal 
line in Fig. 16a until it meets with the second vertical force (2). 
From this point R 2 is then drawn downward to its intersection with 
the horizontal force (30, its intersection with the base of lamina 
(2) giving another point on the line of pressures. This process is 
repeated until the intersection of /? 5 with the final base completes 
the operation for (R.F.). It is evi¬ 
dent that R5 as well as all the other 
resultants are parallel to the cor¬ 
responding ones in Fig. 15, the same 
result being arrived at by different 
graphical processes. 

31. Stepped Polygon. Fig. 

16b is a representation of the so- 
called “stepped” polygon, which 
is also often employed; the form 
differs, but the principle is identical with that already described. 
Inspection of the figure will show that all the resultant lines are 
drawn radiating to one common center or nucleus (0). 

The process of finding the incidence of W on the bases of the 
several lamina is identical with that already described with regard 
to Fig. 15, viz, the same combination of 1+2, 1+2+3, and so on, 
are formed in the funicular 16c and then projected on to the profile. 

32. Modified Equivalent Pressure Area in Inclined Back Dam. 
When the back of a dam is inclined, the area of the triangle of water 
pressure ARC, in Fig. 17, will not equal the product of H, but of Hi 
with its half width, which latter is measured parallel to the base, 
consequently the factor H cannot be eliminated. The triangle 
itself can, however, be altered in outline so that while containing 
the same area, it will also have the vertical height H as a factor 



Fig. 17. Transformation of Inclined 
Pressure Area to Equivalent 
with Horizontal Base 









38 


DAMS AND WEIRS 



Fig. 18. Graphical Method of Fig. 15 Applied to Profile with Curved Back 





















































DAMS AND WEIRS 


39 


in its area. This is effected by the device illustrated in Fig. 17, 
and subsequently repeated in other diagrams. In this figure ABC 
is the triangle of water pressure. By drawing a line CD parallel 
to the back of the wall AB, a point D is obtained on the continu¬ 
ation of the horizontal base line of the dam. A and D are then 
joined. The triangle ABD thus formed is equal to ABC, being 
on the same base AB and between the same parallels. The area 

of ABD is equal to and that of the wall to half width 


BD 

EF X H. Consequently we see that the half width —- of the 


triangle ABD can properly represent the area of the water pressure, 
and the half width EF that of the wall. The vertical height H 
may, therefore, be eliminated. What applies to the whole triangle 
would also apply to any trapezoidal parts of it. The direction of the 
resultant line of water pressure will still be as before, normal to the 
surface of the wall, i.e., parallel to the base BC, and its incidence on 
the back will be at the intersection of a line drawn through the c.g. 
of the area in question, parallel to the base. This point will natur¬ 
ally be the same with regard to the inclined or to the horizontally 
based area. 

33. Curved Back Profiles. In order to illustrate the graphical 
procedure of drawing the line of pressure on a profile having 
a curved back, Figs. 18 and 19 are put forward as illustrations 
merely—not as models of correct design. In these profiles the 
lower two laminas of water pressure, 4' and 5', have inclined 
bases. Both are converted to equivalent areas with horizontal 
bases by the device explained in the last section. Take the lowest 
lamina acdb; in order to convert it into an equivalent trapezoid 
with a horizontal base, de is drawn parallel to ac; the point e is 
joined with A, the apex of the completed triangle, of which the trap¬ 
ezoid is a portion. When af is drawn horizontally, the area acef will 
then be the required converted figure, the horizontally measured 

half width of which multiplied by will equal the area of the original 

o 


trapezoid acdb; — can then be eliminated as a common factor and 
the weights of all the laminas represented in the load line in 




40 


DAMS AND WEIRS 




Fig. 19. Haessler’s Method of Fig. 16 Applied to Curved Back 






















































DAMS AND WEIRS 


41 


Mg. 18a, by the half widths of the several areas. The lamina is 
treated in a similar manner. 

4 he graphical processes in Figs. 18 and 18a are identical with 
those in Fig. 15. In the force polygon 18a the water-pressure forces 
1,2,3, etc., are drawn in directions normal to the adjoining portion 
of the back of the profile on which they abut, and are made equal 
in length to the half widths of the laminas in question. The back 
of the wall is vertical down to the base of lamina 3, consequently 
the forces, 1', 2', and 3 ', will be set out on the water-pressure load 
line in big. ISa from the starting point, horizontally in one line. 
In laminas 4 and 5, however, the back has two inclinations; these 
forces are set out from the termination of 3' at their proper direc¬ 
tions, i.e., parallel to their inclined bases to points marked a and b. 
The direction of the resultants of the combinations, 1' and 2', and 
l r , 2', 3' f will clearly be horizontal. If Aa and Ab be joined, then 
the directions of the combination 1' 2' 3' 4' will be parallel to the 
resultant line Aa and that of 1' 2' 3' 4 ' 5' will be parallel to Ab. 
Thus the inclination of the resultant of any combination of inclined 
forces placed on end, as in the water-pressure load line, will always 
be parallel with a line connecting the terminal of the last of the 
forces in the combination with the origin of the load line. 

34. Treatment for Broken Line Profiles. The method of 
ascertaining the relative position and directions of the resultants 
of water pressure areas when the back of the wall has several inclina¬ 
tions to the vertical is explained as follows: This system involves 
the construction of two additional figures, viz, a force and ray 
polygon built on the water-pressure load line and its reciprocal 
funicular polygon on one side of the profile. These are shown con¬ 
structed, the first on Fig. 18a, the nucleus 0 of the vertical force 
and ray polygon being utilized by drawing rays to the terminations 
of 1', 2', 3', 4', and 5'. In order to construct the reciprocal funic¬ 
ular polygon, Fig. 18c, the first step is to find the c.g.’s of all the 
trapezoidal laminas which make up the water-pressure area, viz, 
1' to 5k This being done, lines are drawn parallel to the bases of 
the laminas (in this case horizontal lines), to intersect the back of 
the wall. From the points thus obtained the force lines 1', 2', 3', 
etc., are drawn at right angles to the portions of the back of the wall 
on which they abut. On these force lines, which are not all parallel, 


42 


DAMS AND WEIRS 


the chord polygon (18c) is constructed as follows: First the initial 
line AO is drawn anywhere parallel to its reciprocal AO, in Fig. 18a. 
From the intersections of this line with the force line F the chord 
marked OF is drawn parallel to OF in Fig. 18a and intersecting 
force line 2'. Again from this point the chord 02' is drawn inter¬ 
secting force 3' whence the chord 03' is continued to force 4', and 
04' up to the force line 5', each parallel to its reciprocal in Fig. 18a. 
The closing line is 05. The intersection of the initial and the closing 
lines of the funicular polygon gives the position of the final resultant 
line V 2' 3' 4' 5', which is then drawn from this point parallel to its 
reciprocal Ob in Fig. 18a to its position on the profile. The other 

resultants are obtained in a sim¬ 
ilar manner by projecting the sev¬ 
eral chords backward till they 
intersect the initial line 0A, these 
intersections being the starting 
points of the other resultants, 
viz, F-4', F-3', F-2', and F. 
These resultant lines are drawn 
parallel to their reciprocals in 
18a, viz, F-4' is parallel to Aa, 
while the remainder are hori¬ 
zontal in direction, the same as 

Fig. 20. Diagram Showing Third Method of their rPPinroooR 
Determining Water Pressure Areas 

This procedure is identical 
with that pursued in forming the funicular 18b, only in this case the 
forces are not all parallel. 

35. Example of Haessler’s Method. In Fig. 19 the profile 
used is similar to Fig. 18, except in the value of p, which is 2{, 
not 2.4 as previously. The graphical system employed is Haessler’s, 
each lamina as already described with reference to Fig. 16 being 
independently dealt with, the combination with the others taking 
place on the profile itself. In this case the changes of batter coin¬ 
cide with the divisions of the laminas, consequently the directions 
of the inclined forces are normal to the position of the back on which 
their areas abut. This involves finding the c. g.’s of each of the water- 
pressure trapezoids, which is not necessary in the first system, unless 
the funicular polygon of inclined forces has to be formed. In spite 






















DAMS AND WEIRS 


43 


of this, in most cases Haessler’s method will be found the handiest 
to employ, particularly in tentative work. 

36. Example of Analytical Treatment. In addition to the 
two systems already described, there is yet another corresponding 
to the analytical, an illustration of which is given in Fig. 20. In 
this the vertical and horizontal components of R, the resultants 
(R.F.), viz, N and P are found. In this method the vertical com¬ 
ponent of the inclined water pressure Pi is added to the vertical 
weight of the dam itself, and when areas are used to represent 
weights the area of this water overlying the back slope will have to 
be reduced to a masonry base by division by the specific gravity of 
the masonry. 

37. Relations of R. N. and W. The diagram in Fig. 20 is a 
further illustration showing the relative positions of R, P, Pi, N, 
and IF. The line R starts from a, the intersection of the horizontal 
force P with N, the resultant of all the vertical forces, for the reason 
that it is the resultant of the combination of these two forces; but 
R is also the resultant of Pi and IF, consequently it will pass through 
a', the intersection of these latter forces. The points a and cq are 
consequently in the resultant R and it follows as well that if the 
position of R is known, that of N and IF can be obtained graphically 
by the intersection of P or Pi with P. These lines have already 
been discussed. 


UNUSUALLY HIGH DAMS 

38. “High” Dams. An example will now be given, Fig. 21, of 
the design of a high dam, i.e., one whose height exceeds the limit 
before stated. As usual the elementary triangular profile forms 
the guide in the design of the upper portion. We have seen in 
section 24 that the limiting depth with p = 2.4 and c = 16 tons = 
195 feet, whence for 18 tons’ limit the depth will be 219 feet. In 
Fig. 21 the tentative profile is taken down to a depth of 180 feet. 
The crest is made 15 feet wide and the back is battered 1 in 30; the 
base width is made 180X.645 = 116 feet. The heel projects 6 feet 
outside the axis line. The graphical procedure requires no special 
explanation. It follows the analytical in dealing with the water 
pressure as a horizontal force, the weight of the water overlying 
the back being added to that of the solid dam. For purposes of 




44 


DAMS AND WEIRS 



a 

c3 

Q 

Id 

bC 

s 


aj 

bo 

<3 


o3 

o 

• 

"c3 

£ 

*< 


<N 

M 

• pH 





















































DAMS AND WEIRS 


45 


calculation the load is divided into three parts (1) the water on 
the sloping back, the area of which is 540 sq. ft. This has to be 
reduced by dividing it by p and so becomes 225 sq. ft. As tons, 
not areas, will be used, this procedure is not necessary, but is adopted 
for the sake of uniformity in treatment to avoid errors. The c.g. 
of (1) is clearly 2 feet distant from the heel of the base of (3), about 
which point moments will be taken. That of the crest (2) is 13.3 feet 
and that of the main body (3) obtained by using formula (7) comes 
to 41.2 feet. The statement of moments is then as follows: 


No. 

Area 

Tons 

Lever Arm 

Moment 

1 

225 

17 

2 

34 

2 

360 

27 

13.3 

359 

3 

10280 

771 

41.2 

31765 

Total 

10866 

815 


32158 


Then the distance of N from the heel will be '. ) ^ = 39.4 ft.; it thus 

815 

falls 39.4 —38.7 = .7 ft. within the middle third. The distance (/) 

p a 

between N and R is — rp Now P = the area of the right angle 

o iV 

triangle whose base is or 75 feet, and is 6750 sq. ft. equivalent 

P 

, 6750X3 _ , . , 506X180 

to-= 506 tons. I he expression then becomes j = - - - 

40 oXoio 

= 37.2 feet. The incidence of R will then be 37.2+39.4 = 76.6 feet 


distant from the heel of the base. The — point being 77.3 from the 

O 

heel, R falls .7 ft. within. Thus far the tentative profile has proved 
fairly satisfactory, although a slight reduction in the base width is 
possible. The position of W, or the resultant weight of the portions 
2 and 3 of the dam is obtained from the moment table already 
given, and is the sum of the moments of (2) and (3) divided by (2) 

i (3) or 3212 i = 40.2 ft. This falls 40.2-38.7 = 1.5 ft. within the 
w 798 

























46 


DAMS AND WEIRS 


middle third. The value of q (R.F.) is— — .7—19.33 —.7 — 18.6 

o 

feet, and m>= ( 1 = 1 +~T77T = 1 -96. 

V b / lib 

Then by formula (10i), N being 815 and P, 506 

1.96x815+V(1.96x815) 2 + 4 (506) 2 
C 232 _ 

1597+^2551367+1024144 

232 


= 15.03 tons per square foot 

Extension of Profile. This value being well below the limit 
of 18 tons and both resultants (R.F.) and (R.E.) standing within the 
middle third it is deemed that the same profile can be carried down 
another 30 ft. in depth without widening. The base length will 
now be a trifle over 135 feet. The area of the new portion (4) is 
3769 sq. ft. =283 tons. The distance of its c.g. from the heel by 
formula (7) is found to be 63.4 feet. The position of W will be 
obtained as follows, the center of moments being one foot farther 
to right than in last paragraph. 


No. 

Tons 

Lever Arm 

Moment 

2 

27 

14.3 

386 

3 

771 

42.2 

32536 

4 

283 

63.4 

17942 

Total 

1081 


50864 

. 


.*. x = = 47.0 feet from heel of new base to W 

1081 

1 1 QP 

As is -—!■ =45 ft., the incidence of W\ is 2.0 ft. within the base, 

o o 


which is satisfactory. 

To find Nit the moment of the water on whole back can be 
added to that of W first obtained. The offset from the axis being 

210X7 

now 7 ft., the area will be —-— = 735 equal to an area of masonry 


of 


735 

2.4 


= 306 sq. ft. equivalent to 23 tons, nearly. The lever arm 


























DAMS AND WEIRS 


47 


being y, or 2.3 feet the moment about the heel will be 23X2.3 = 52.9, 


say 53.0 ft.-tons. This amount added to the moment of W\ will 
represent that of Ah and will be 50,864+53 = 50,917. The value of 
Ah is that of RT+ the water on back or, 1081+23 = 1104 tons. 


The distance of Ah from the heel is then ^^- 7 = 46.2 feet. To 

1104 

obtain that of 7?i the value of /== 43.7 feet; this added 

3 X1104 

to 46.2 = 89.9 ft., the incidence of Ri is therefore — w — 89.9 = 


90—'89.9 —.1 ft., within the middle third boundary. Then q = — 

6 

— .1 = 22.5 —.1 = 22.4 ft., and m= 1+— = 1+"T^~ = 2.00, nearly. 

b 135 

To find c, formula (10) will be used, the quantities being less than 

mN 2.00X1104 


in formula (10i). Here s = 


J P_689 
b “135 


135 


= 16.4 tons and s s 


= 5.1 tons. Then 


16.4 , 

c= ~f+\ 


(16.4) : 


■f (5.1) 2 = 8.2+97 = 17.9 tons 


The limit of 18 tons, being now reached, this profile will have to 
be departed from. 

39. Pentagonal Profile to Be Widened. The method now to 
be adopted is purely tentative and graphic construction will be 
found a great aid to its solution. A lamina of a depth of 60 ft., 
will be added to the profile. It is evident that its base width must 
be greater than that which would be formed by the profile being 
continued down straight to this level. The back batter naturally 
will be greater than the fore. From examination of other profiles it 
appears that the rear batter varies roughly from about 1 in 5 to 1 
in 8 while the fore batter is about 1 to 1. As a first trial an 8 ft. 
extra offset at the back was assumed with a base of 200 feet; this 
would give the required front projection. Graphical trial lines 
showed that N would fall without the middle third, and W as well; 
the stress also just exceeded 18 tons (R.E.). A second trial was 
now made in which the back batter was increased and the base 
shortened to 180 feet. In this case c exceeded the 18 ton limit. 













48 


DAMS AND WEIRS 


Still further widening was evidently required at the heel in ordei 
to increase the weight of the overlying water, while it was clear 
that the base width would not bear reduction. The rear offset 
was then increased to 15 feet and the base width to 200 feet. The 
stresses now worked out about right and the resultants both fell 
within the middle third. By using formula (6) the distance of the 
c.g. of the trapezoid of water pressure, which weighs 112 tons, was 
found to be 7.2 feet from the heel of the base, and by formula (7) 
that of the lowest lamina (5) from the same point is found to be 
91.8 feet; the weight of this portion is 754 tons. These two new 
vertical forces can now be combined with N i whose area and posi¬ 
tion are known and thus that of iV 2 can be ascertained. Ni is 46.2 
ft. distant from the heel of the upper profile; its lever arm will, 
therefore, be 46.2 + 15 = 61.2 feet. The combined moment about 
the heel will then be 



Weight 

Lever Arm 

Moment 

Water 

112 

7.2 

806 

N i 

1104 

61.2 

67565 

(5) 

754 

91.8 

69217 

Total 

1970 


137588 


I 

The incidence of iV 2 is then H =70 feet from the heel; as the 


1970 


200 


middle third boundary is = 66.6 feet distant from the same point, 

O 

N 2 falls 3.4 feet within. The distance between A T 2 and R 2 (viz, /) 
PH 

= ——. Now P 2 , or the horizontal water pressure, has a reduced area 
3 N 2 

1139 X 9 70 

of 15187 feet, equivalent to 1139 tons, consequently /= t) ^ ^7^ - 

= 52.0 feet. This fixes the incidence of R 2 at 70+52.0 = 122.0 feet 
distant from the heel; the § point is 133.3 feet distant, consequently 

R 2 falls well within the middle third and as q= 122.0 — :: ^ = 22.0 feet, 

,, 6X22.0 , , mN 1.66X1970 0 , 

m— H—— = 1-66, and s =— — =--=16.3 tons. 


200 


200 





















DAMS AND WEIRS 


49 


Now 


S.Q 


P 2 1139 
b ~ 200 


= 5.7 tons 


Whence by formula (10), 


16.3 1 205 

c— 2 +32.5 = 8.15+9.9 = 18.05 tons 

which is the exact limit stress. 

The value of s 2 (the pressure at the heel) is obtained by the 

130 S 

same formula, using the minus sign, viz, m = 1—1—— = .34, there- 

200 

fore, s 2 =—r^ = —3.4 tons, nearly. These vertical reac- 

b 200 

tions are set out below the profile. With regard to W 2 it is com¬ 
posed of 1 Ei+( 5). The table of moments is as follows: 



Weight 

Lever Arm 

Moment 

w , 

1081 

62.0 

67022 

(5) 

754 

91.8 

69217 

W 2 

1835 


136239 


The distance of W 2 from the heel is 


136239 

1835 


falls 7.6 feet within the middle third; q = 


= 74.3. W 2 , therefore, 
200 

—74.3 = 25.7 feet, 

O * 


a v 9 p; 7 

and m = 1 + -——- 1 - = 1.77, and m = 1 — .77 = .23; therefore, s = 

200 

mW 2 1.77X1835 
b ~ 200 


1835 

= 16.2 tons, and ^ 2 = .23X-^—j =2.1 tons. These 


pressures are shown below the profile. 

The value of 6 in all three cases is less than 35° which is also 

one of the stipulations. 

In continuing the profile below the 270-foot depth the proba¬ 
bility is that for a further depth of 50 or 60 feet the same fore and 
rear batter would answer; if not, the adjustment is not a difficult 
matter to manipulate. As previously stated, the incidence of A 
should be fixed a little within the middle third when that of W and 
R will generally be satisfactory. 





























50 


DAMS AND WEIRS 


In the force diagram the water part of N is kept on the top of 
the load line W. This enables the lengths of the N series to be 
clearly shown. The effect is the same as if inclined water pressure 
lines were drawn, as has already been exhibited in several cases. 

40. Silt against Base of Dam. In Fig. 21, suppose that the 
water below the 210-foot depth was so mixed with silt as to have a 
specific gravity of 1.4 instead of unity. The effect of this can be 
shown graphically without alteration of the existing work. In the 
trapezoid lying between 210 and 270 the rectangle on ab represents 
the pressure above 210 and the remaining triangle that of the lower 

60 feet of water. The base of the latter, be is, therefore, — = 7T7 = 


25 feet. Now the weight of the water is increased in the proportion 


of 1.4 : 1, consequently the proper base width will be 


H'X 1.4 
2.4 


60X1.4 

2.4 


= 35 feet. 


The triangle acd then represents the additional 


pressure area due to silt. The normal pressure on the back of the 
dam due to the presence of silt is shown graphically by the triangle 
attached, whose base = cd=10 feet; its area is 310 square feet, 
equivalent to 23 tons. This inclined force is combined with R 2 at 
the top right-hand corner of Fig. 21a and the resultant is R 3 ; on the 
profile the reciprocal inclined force is run out to meet R 2 and from 
this intersection R3 can be drawn up toward P 2 . This latter inter¬ 
section gives the altered position of N 2 , which is too slight to be 
noticeable on this scale. The value of c and the inclination of R 
are both increased, which is detrimental. 

If the mud became consolidated into a water-tight mass the 
pressure on the dam would be relieved to some extent, as the earth 
will not exert liquid pressure against the back. Liquid mud pres¬ 
sure at the bottom of a reservoir can consequently be generally 
neglected in design. 

41. Filling against Toe of Dam. Now let the other side of the 
dam be considered. Supposing a mass of porous material having an 
immersed s. g. of 1.8 is deposited on the toe, as is often actually the 


case. 


Then a pressure triangle of which the base equals H X 


1.8 

2.4 


45 feet is drawn; its area will be 1755 and weight 132 tons; the 






DAMS AND WEIRS 


51 


resultant P 4 acting through its c. g. is run out to intersect P 2 . At the 
same time from the lower extremity of P 2 , in the force diagram, a 
reciprocal pressure line P 4 is drawn in the same direction equal in 
length 132 tons and its extremity is joined with that of P 2 ; the result¬ 
ing line 1U is then projected on the profile from the previous inter¬ 
section until it cuts the force line P 2 ; this gives a new resultant P 4 
and a new position for N, viz, iV 4 , which is drawn on the profile; JV also 
will be similarly affected. The load on the toe of the dam increases 
its stability as the value of 6 is lessened, the position of W is also 
improved, but that of P 4 , which is nearer to the toe than P 3 , is not. 
To adjust matters, the c. g. of (5) requires moving to the right which 



is affected by shifting the base line thus increasing the back and 
decreasing the front batter, retaining the base length the same as 
before. 

42. Ice Pressure. Ice pressure against the back of a dam 
has sometimes to be allowed for in the design of the profile; as a 
rule, however, most reservoirs are not full in winter so that the 
expansive pressure is exerted not at the summit but at some distance 
lower down where the effect is negligible. In addition to this when 
the sides of a reservoir are sloping, as is generally the case, movement 
of ice can take place and so the dam is relieved from any pressure. 
In the estimates for the Quaker Bridge dam it is stated that an 
ice pressure of over 20 tons per square foot was provided for. No 
















52 


DAMS AND WEIRS 


definite rules seem to be available as to what allowance is suitable. 
Many authorities neglect it altogether. 

The effect of a pressure of ten tons per foot run on a hundred- 
foot dam acting at the water level is illustrated in Fig. 22. For 
this purpose a trapezoidal section has been adopted below the sum¬ 
mit level. The crest is made 15 feet wide and 10 feet high. This 
solid section is only just sufficient, as will appear from the incidence 
of R' on the base. The area of this profile is 4150 sq. ft., while one 
of the ordinary pentagonal sections as dotted on the drawing would 



contain but 3325 sq. ft. The increase due to the ice pressure is 
therefore 825 sq. ft. or about 25 per cent. The graphical procedure 
hardly needs explanation. The ice pressure p is first combined 
with IF the weight of the dam and their resultant R cuts P at a 
point from which the final 7?i is run down to the base parallel to its 
reciprocal in the force diagram. It falls just within the middle 
third of the base. An actual example is given in section 56. 

43. Partial Overfall Dams. It not infrequently happens that 
the crest of a dam is lowered for a certain length, this portion acting as 
a waste weir, the crest of the balance of the dam being raised above 
the water level. In such cases a trapezoidal outline is generally 
preferable for the weir portion and the section can be continued 
upon the same lines to form the upper part of the dam, or the upper 
part can be a vertical crest resting on the trapezoidal body. In a 


































DAMS AND WEIRS 


53 


trapezoidal dam, if the ratio of 



be r, the correct base width is 


obtained by the following formula: 





This assumes the crest and summit water level to be the same. 
In Fig. 23, p is taken as 2.4 and r as .2. The base width with a 

vertical back will then be X — = ^ ■ = 50 X .645 X .935 = 31.3 

Vp Vl-f-.2-.04 

feet, and the crest width lc will be 31.3 X .2 = 6.3 feet. In the second 
figure the profile is shown canted forward, which is desirable in 
weirs, and any loss in stability is generally more than compensated 
for by the influence of the reverse pressure of the tail water which 
influence increases with the steepness of the fore slope of the weir. 
The base width is, however, increased by one foot in the second figure. 
As will be seen in the next section, the crest width of a weir 

should not be less than V/7 + Vd; in this case 77 = 45 and d = 5. 
This would provide a crest width of 6.7 + 2.2 = 9 feet, which it nearly 
scales. 

NOTABLE EXISTING DAMS 

44. Cheeseman Lake Dam. Some actual examples of dam 
sections will now be exhibited and analyzed. Fig. 24 is the section 
of the Cheeseman Lake dam near Denver, Colorado, which is one 
of the highest in the world. It is built to a curvature of 400 feet 
radius across a narrow canyon. It is considered a gravity dam, 
however, and will be analyzed as such. The section can be divided 
into three unequal parts 1, 2, and 3, and the lines of pressure (R.F.) 
and (R.E.) will be drawn through the bases of these three divisions. 
Of the vertical forces (1) has an area of 756 sq. ft., (2) of 3840, and 
(3) of 13,356 the total value of W being 17,952 sq. ft., which is marked 
off on the load line in Fig. 24a. With regard to the water-pressure 
areas the most convenient method, where half widths are not used, 
which can only be done with equal divisions, is to estimate the 
areas of the horizontal pressures only and set them off horizontally, 
the values of the inclined pressures being obtained by construction. 
For this purpose the triangle of horizontal water pressure is shown 
adjacent to, but separate from, the profile. The three values of P 








54 


DAMS AND WEIRS 


which are equal to 


IP 

2 p 


will be 270, 2631, and 7636, respectively, the 


total being 10,537 sq. ft. In this computation the value of p is 
assumed to be 2.4. These several lengths are now set out hori¬ 
zontally from the origin 0 in Fig. 24a, and verticals drawn upward 
intercept the chords, 1', 2', 3', which latter are drawn from the 
origin 0, parallel to their respective directions, i.e., normal to the 
adjacent parts of the wall. The rest of the process is similar to that 
already described, with reference to Figs. 16 and 18, and need not 
be repeated. In Fig. 24a N scales 19,450, equal to 1457 tons, and 



i r-, rp, p mN 1.51X1457 10 _ . P 783 

1.51. therefore, s = —— =-—-= 12.5 tons, and s s =y=—= 

4.45; then by formula (10) 

12.5 . 1(12.5)' 


c — 


!.5 , 1(12. 

r+\-j 


1-(4.45) 2 = 6.25+V59 = 13.9 tons, approx. 


With regard to IF, q scales about 20 ft., m then works out to 1.7, 

, , mW 1.7X1346 10 _ 

nearly, and s=—j— =-—-— = 13.0 tons. 

b 176 

As an exercise the inclined final resultant P is drawn on the 
profile. Ihis line is parallel to Oc in Fig. 24a, its location is worked 
out by means of the funicular polygon, the construction of which 
need not be explained after what has gone before. 

















































DAMS AND WEIRS 


55 


45. Analytical Check. In order to check this result analyt¬ 
ically the procedure will be, first, calculate the position of the c. g. of 
the trapezoids (2) and (3) relative to the rear corner of their bases 
by formula (7) and also the positions of the resultants of the vertical 
components of the water pressure overlying the back with regard 
to the same points by formula ( 6 ). Second, convert the areas into 
tons by multiplying by ^o. The statement of moments about the 
heel of the base, w T ith the object of finding the position of W is given 
below. 


Moment of ( 1 ) 

56.7 X32.5 

= 1843 

Moment of ( 2 ) 

288 X47.9 

= 13795 

Moment of (3) 

1001 X75 

= 75075 

Total W = 

1346 tons 

90713 

The distance of W from the heel will then be 

90713 A7r 

1346 =67 ‘ 5 

order to obtain N, the moments of the water weights will 

be added as below. 



Moment of W 

1346X67.5 

= 90713 

Moment of Wi 

10X21.6 

= 216 

Moment of w 2 

107X9 

= 963 

Total N = 

1463 tons 

91892 


and 


91892 _ 2 g £ ^ 


To find the incidence of R and its distance ( q ) from the center 
point, that from the known position of N must be computed from 

PH 783X224 


the formula /= 


176 


3 N 3X1463 


= 40 ft., therefore, q= (62.8+40.0) 


= 14.8 feet. This is close to the value obtained graphically 

z 

which was taken as 15 feet. The value of N is also seen to be close 
to that obtained graphically. The value of q with regard to W (R.E.) 


is as follows, <7 = — - 67.5 = 20.5 feet, almost exactly what it scales on 

2 
























56 


DAMS AND WEIRS 




the diagram. In this profile the upper part is light, necessarily made 
up for in the lower part. 

At the upper base line of (2) the incidence of W is exactly at 
the middle third edge, while R falls within it. At the final base the 
position of N is 62.8 distant from the heel and the inner third point 


176 

is —- = 58.6 distant, consequently the incidence of N lies 4.2 feet 

O 


within the boundary. 

If the position of N were made obligatory at the inner edge of 
the middle third, the value of W would be increased, but R would 



be decreased. There may have been special reasons for limiting the 
maximum stress (R.E.). On Fig. 24 the position of N is obtained 
by the intersection of the horizontal resultant P with R prolonged 
upward. If the stress were calculated on the supposition that the 
structure was an arched dam, it would amount to 21J tons by the 
“long” formula, given in section 78, Part II. 

46. Roosevelt Dam. In Fig. 25 is given the profile of the 
Roosevelt dam, and Fig. 26 is the site plan of that celebrated work. 
For some years, the Roosevelt dam was the highest gravity dam in 
existence. It spans a very deep canyon of the Salt River in Arizona 















































DAMS AND WEIRS 


57 


and impounds the enormous quantity of 1 J million acre-feet of water, 
which will be utilized for irrigation. This work is part of one of 
the greatest of the several large land reclamation projects under¬ 
taken by the U. S. Government for the watering and settling of 
arid tracts in the dry zone of the western states. 

The profile is remarkable for the severe simplicity of its out¬ 
line. It closely follows the elementary profile right down to its 
extreme base and forms a powerful advocate for this simple style 
of design. The graphical procedure is similar to that in the last 
example. The section is divided into three divisions. As the 
first two are comparatively small, the triangle of forces in Fig. 25a 



is first plotted at a large scale in pencil and the inclinations of the 
resultants thus obtained are transferred to the profile; this accounts 
for the long projecting lines near the origin of the force diagram 
which also appear in some previous examples. A neater method 
for overcoming this difficulty is that adopted in the next figure, 
when the forces ( 1 ) and ( 2 ) are first amalgamated into one before 
being plotted on the force diagram. 

In Fig. 25a, N scales roughly 19,000 sq. ft., equivalent to 1425 


120 

tons, q also measures approximately 20 ft., then m = 1 + l 6 T L75 ’ 

s s = — = —=5.1 tons. By 


and s = 


N 1425X1.75 


160 


= 15.5 tons. 


P _ 826 
= 160 


formula ( 10 ) 





























58 


DAMS AND WEIRS 




(5.1 ) 2 = 7.75 + V86 = 17 tons roughly 


With regard to W, q measures 23 ft. and m works out to 1.86 

,, , mW 1.86X1378 

therefore s =—j— = -- =16 tons per sq. it. 

b 160 


This dam is built on a radius of 410 feet, measured from the 
axis; if measured from the extrados of the curve at the base it will 
be 420 feet and the arch stress as calculated from the “long” formula 
used in “Arched Dams” will amount to 23.3 tons. 

The site plan given in Fig. 26 forms an instructive example of the 
arrangement of spillways cut in the solid rock out of the shoulders 
of the side of the canyon, the material thus obtained being used in 
the dam. These spillways are each 200 feet wide and are excavated 
down to five feet below the crest of the dwarf waste weir walls which 
cross them. This allows of a much greater discharge passing under 
a given head than would be the case with a simple channel without 
a drop wall and with bed at the weir crest level. The heading up, 
or afflux, is by this means diminished and that is a matter affecting 
the height given to the dam crest. 

47. New Croton Dam. The profile of the New Croton dam 
constructed in connection with the water supply of New York City 
is given in Fig. 27. This dam has a straight alignment and is 1168 
feet long. Waste flood water is accommodated by an overfull weir 
1000 ft. in length, which is situated on one flank forming a continua¬ 
tion of the dam at right angles to its axis. The surplus water 
falls into the Rocky River bed and is conveyed away by a separate 
channel. An elevation and plan of this work are given in Figs. 
28 and 29. 

The system of graphical analysis employed in this case is differ¬ 
ent from that in the last two examples and is that illustrated in Fig. 
18, where independent combinations of vertical and inclined forces 
are used. The profile is divided into four divisions, the first being 
a combination of two small upper ones. The further procedure after 
the long explanations already given does not require any special 
notice except to point out that the directions of the combined forces 
1 /, F+2', F+2'4-3' etc., in ( d ) are drawn parallel to their reciprocal 
lines on Fig. 27a, namely to the chords Oa, Ob, Oc, and Od, respec¬ 
tively. The final resultants are i? 4 (R.F.) and W (R.E.). The 







DAMS AND WEIRS . 


59 


value of II is 1380 tons and that of N is 1484 tons, consequently 
applying formula (10), q in the first case scales 26 feet and m works 

out to 1.82 therefore s =—— =-—-=13.2 tons =c, as with 


190 


W, s and c are identical. 


With regard to N, q scales 7 feet, consequently m = Id- 


42 


190 

i 92v1484 

1.22, s= - 77 ^-= 9.5 tons only. As P = 10,010 = 750 tons, 


190 


P 


9.5 


— = 4 tons; therefore c — 
b 2 


f(9.5) : 


-f (4) 2 = 11 tons, which is very 


moderate. It is probable that other external pressures exist due 



Fig. 27. Diagram of Profile of New Croton Dam Showing Influence Lines as in Fig. 18 


to filling in front and rear, as also ice pressure, which would materi¬ 
ally modify the result above shown. This dam, like the Cheese- 
man, is of the bottleneck profile, it is straight and not curved on plan. 

48. Assuan Dam. The section, Fig. 30, is of the Assuan dam 
in Egypt, which notable work was built across the Nile River above 
the first cataract. As it stands at present it is not remarkable for 































































WEIR CREST R L E77.0 





Fie. 29. Plan of New Croton Dam for New York City Water Supply 
























































































































DAMS AND WEIRS 


G1 


its height, but what it lacks in that respect, as in most eastern 
works, is made up in length, which latter is 6400 feet. No single 
irrigation work of modern times has been more useful or far-reaching 
in beneficial results upon the industrial welfare of the people than 
this dam. Its original capacity was 863,000 acre-feet and the back 
water extended for 140 miles up the river. The work is principally 
remarkable as being the only solid dam which passes the whole 
discharge of a large river like the Nile, estimated at 500,000 second- 
feet, through its body, for which purpose it is provided with 140 
low and 40 high sluices. These are arranged in groups of ten, each 



Fig. 30. Assuan Dam across the Nile Showing Old and New Profiles 


low sluice is 23 feet deep by 6J feet wide with the dividing piers 
16§ feet wide. The diminution of the weight of the dam due to 
sluices necessitates an excess of width over what would be sufficient 
for a solid dam; in addition to which the maximum pressure in the 
piers is limited to the extremely low figure of 5 tons of 2000 pounds. 
The designers have thus certainly not erred on the side of boldness; 
the foundation being solid granite, would presumably stand, with 
perfect safety, pressure of treble that intensity, while the masonry, 
being also granite, set in cement mortar, is certainly capable of 
carrying a safe pressure of 15 tons, as many examples prove. 




































EASTERN DESERT HILLS 

V : ;r 

jimrnm 



^ Is. ®Sl 
* 

<Q < Si 


& JO C> 

2> <M 

*> ^ T> 


:, n HHI/h///^ . 


''nii'iiuhiiitillhv^^."‘J{ 

WESTERN DESERT ' H/LIS 



Fig. 31. Location Plan and Longitudinal Section of Assuan Dam 
























































































































































DAMS AND WEIRS 


63 


This dam has proved such a financial success that it has recently 

1/ 

been raised by 23 feet to the height originally projected. The 
water thus impounded is nearly doubled in quantity, i.e., to 
over 1| million acre-feet; exceeding even that of the Salt River 
reservoir in Arizona. As it was decided not to exceed the low unit 
pressure previously adopted, the profile has been widened by 16§ 
feet throughout. A space has been left between the new and the 
old work which has been subsequently filled in with cement grout 
under pressure, in addition to which a series of steel rods has been let 



Fig. 32. View of Assuan Dam before Being Heightened with Sluices in Operation 


into the old face by boring, and built into the new work. The enlarge¬ 
ment is shown in the figure. The sluices are capable of discharging 
500,000 second feet; as their combined area is 25,000 square feet 
this will mean a velocity of 20 feet per second. Owing, however, to 
the possibility of adjustment of level, by manipulation of the sluice 
gates, they will never be put to so severe a test. 

A location plan and longitudinal section shown in Fig. 31, a 
view of the sluices in operation, Fig. 32, and a view of the new work 
in process, Fig. 33, will give a good idea of the construction features. 




64 


DAMS AND WEIRS 



Fig. 33. View of Assuan Dam Showing New Work in Progress 







DAMS AND WEIRS 


65 


49. Cross River and Ashokan Dams. Two further sections are 
given in Figs. 34 and 35, the first of the Cross River dam, and the 
second of the Ashokan dam in New York. Both are of unusually 
thick dimensions near the crest, this being specially provided to 
enable the dams to resist the impact of floating ice. These profiles 
are left to be analyzed by the student. The Ashokan dam is pro¬ 
vided with a vertical line of porous blocks connected with two inspec¬ 
tion galleries. This is a German innovation, which enables any 



leakage through the wall to be drained off, thereby guarding against 
hydrostatic uplift. This refinement is now frequently adopted. 

50. Burrin Juick Dam. The Burrin Juick dam in Australia, 
Fig. 36, which is generally termed “Barren Jack”, is a close copy of 
the Roosevelt dam, Fig. 25, and is a further corroboration of the 
excellence of that profile. It is built across the Murrumbidgee 
River in New South Wales not far from the new Federal Capital. 
Its length is 784 feet on the crest, the maximum height being 240 
feet. The fore batter is 3 vertical to 2 horizontal, and the back 
batter 20 vertical to 1 horizontal, both identical with those adopted 
in the Roosevelt dam; the crest width is 18 feet. It is built on a 
curve to a radius of 1200 feet. This dam will impound 785,000 
acre-feet. The material of which the dam is composed is crushed 
sandstone in cement mortar with a plentiful sprinkling of large 
“plums” of granite. The ultimate resistance of specimen cubes 
































66 


DAMS AND WEIRS 


was 180 “long” tons, per square foot; the high factor of safety of 
12 was adopted, the usual being 8 to 10. The maximum allowable 
stress will, therefore, reduce to 15 “long” tons = 16.8 American 
short tons. 

With regard to the maximum stresses, for Reservoir Full , N = 
16,100, equivalent to 1210 tons, and q scales about 15 feet, conse- 



9200 


FAD. 1200 


ASSUMED FULL 
SUPPLY LEVEL 


A=I47I8 * 


15580 




F^—145'A -*- 


Fig. 36. Analytical Diagram Showing Profile of Burrin Juick Dam in Australia 


i i 1.62X1210 10 r i 

quently m comes to 1.62, and s = —-— =-- 13.5 tons, and 


145 


P _ 690 _ ^ g tons. Whence 
b 145 


Si , Is 2 , 9 13.5, l( 13.5) 2 . OX9 

c = y+^| — +Ss ~2 — 4 - + (4.8) 2 = 15 tons 

For Reservoir Empty, W — 15,580 feet or 1170 tons, g = 24, m = 2 

mW 2X1170 


s = 


145 


= 16 tons, nearly 





















































DAMS AND WEIRS 


67 


The above proves that the stress (R.E.) is greater than that of 
(R-F.). Probably allowance was made for masses of porous filling 
lying at the rear of the dam, which would cause N and W to be 
shifted forward and so equalize the pressure. It will be noticed 
that the incidence of N, the vertical component (R.E.) falls exactly 
at the edge of the middle third, a condition evidently observed in 
the design of the base width. 



Fig. 37. Profile of Arrow Rock Dam, Idaho, Showing Incidence of Centers of Pressure on Base 


The dam is provided with two by-washes 400 feet wide; the 
reservoir will be tapped by a tunnel 14X13 feet, the entrance 
sluices of which will be worked from a valve tower upstream, a 
similar arrangement to that in the Roosevelt dam. It is interesting 
to note that an American engineer has been put in charge of the 
construction of this immense work by the Commonwealth Govern¬ 
ment. 

51. Arrow Rock Dam. The highest dam in the world now 
just completed (1915) is the Arrow Rock on the Boise, Idaho, 
project, a U. S. reclamation work. From the crest to the base the 
fore curtain is 351 feet. A graphical analysis of the stress in the 



























68 


DAMS AND WEIRS 


base is given in Fig. 37. For Reservoir Empty , If r = 2609 tons, and 

228 2 IF 

q measures 38 feet; therefore m= 1 + ^33 = 2 nearly and s = —y~ = 

0 2609 no _ , T , „ • r 77 o-t i ii 

2X— = 23.5 tons, kor Reservoir Full , q = 2i, and w=l+—— = 
222 zzz 

1.73X2609 OAO , P 1610 7Q . , 20.2 

-—-= 20.2 tons. = y = —— = 7.3 tons, and c = —+ 


\ 


( 20 . 2 ) 


( 7 . 3) 2 = 22.6 tons. These values are, of course, but 


approximate. 

Thus the compressive stresses (R.F.) and (R.E.) are practically 
equal, and the incidence of IF and also of N is close to the edge of 



Fig. 38. Location Plan of Arrow Rock Dam 
Courtesy of "Engineering Record ” 


the middle third. The dam is built on a radius of 661 feet at the 
crest. The high stresses allowed are remarkable, as the design is on 
the gravity principle, arch action being ignored. The curvature 
doubtless adds considerably to safety and undoubtedly tends to 
reduce the compressive stresses by an indeterminate but substantial 
amount. It is evident that formula ( 10 ) has been applied to the 




















DAMS AND WEIRS 


G9 


design. Reference to* Figs. 38 and 39 will show that the dam is 
divided into several vertical sections by contraction joints. It is 
also provided with inspection galleries in the interior and vertical 
weeper drains 10 feet apart. These intercept any possible seepage, 
which is carried to a sump and pumped out. These precautions are 



to guard against hydrostatic uplift. The simplicity of the outline, 
resembling that of the Roosevelt dam, is remarkable. 

SPECIAL FOUNDATIONS 

52. Dams Not Always on Rock. Dams are not always founded 
on impervious rock but sometimes, when of low height, are founded 
on boulders, gravel, or sand. These materials when restrained 
from spreading, and with proper arrangements to take care of sub- 
percolation, are superior to clay, which latter is always a treacher¬ 
ous material to deal with. When water penetrates underneath the 
base of a dam, it causes hydrostatic uplift, which materially reduces 
the effective weight of the structure. Fig. 40 represents a wall 
resting on a pervious stratum and upholding water. The water 
has ingress into the substratum and the upward pressure it will 
exert at c against the base of the wall will be that due to its depth, 
in this case 30 feet. Now the point of egress of the percolation will 
be at b, and, as in the case of a pipe discharging in the open, pressure 
is nil at that point; consequently the uplift area below the base will 

be a triangle whose area equals The diagram, Fig. 40, shows 

JmJ 

the combinations of the horizontal water pressure P with the hydro¬ 
static uplift V and with the weight of the wall IF. P is first com¬ 
bined with V, Ri resulting, whose direction is upward. i?i is then 


























70 


DAMS AND WEIRS 


combined with W, R 2 being their resultant. The conditions with¬ 
out uplift are also shown by the dotted line drawn parallel to dc in 
Fig. 40. The line ab is termed the hydraulic gradient; it is also the 
piezometric line, i.e., a line connecting water levels in piezometer 
tubes, were such inserted. 

Fig. 41 shows the same result produced on the assumption that 
the portion of the wall situated below the piezometric line is reduced 
in weight by an equal volume of water, i.e., the s.g. of this part may 
be assumed reduced by unity, i.e., from 2.4 to 1.4. The wall is 



thus divided diagonally into two parts, one of s.g. 2.4 and the other 
of s.g. 1.4. The combination of 1+2 with P is identical in result 
with that shown in Fig. 40. In the subsequent section, dealing 
with “Submerged Weirs on Sand”, this matter of reduction in weight 
due to flotation is frequently referred to. 

53. Aprons Affect Uplift. Fig. 42 is further illustrative of the 
principle involved in dams with porous foundations. The pentag¬ 
onal profile abc, is of sufficient base width, provided hydrostatic 
uplift is absent. Supposing the foundation to be porous, the area 






























































DAMS AND WEIRS 


71 


of uplift will be cq be, in which 6oq, equals ab. This area is equal 
to abc, consequently practically the whole of the profile lies below 
the hydraulic gradient, may be considered as submerged, and hence 
loses weight; its s.g. can thus be assumed as reduced by unity, i.e., 
from p to p — 1. The correct base width will then be found by 

making b = - JL= instead of The new profile will then be adb; 

Vp —1 vp 

the base width having been thus extended, the uplift is likewise 
increased in the same proportion. Now supposing an impervious 
apron to be built in front of the toe as must be the case with an 
overfall dam; then the area of uplift becomes barf, and the piezo- 



Fig. 41. Diagram Showing Identical Result If Weight Is Considered Reduced 

Due to Submersion 


metric line and hydraulic gradient, which in all these cases happen 
to be one and the same line, is ae. Under these circumstances the 
comparatively thin apron is subjected to very considerable uplift 
and will blow up unless sufficiently thick to resist the hydrostatic 
pressure. The low water, or free outlet level, is assumed to be at 
the level e, consequently the fore apron lies above this level and is 
considered as free from flotation due to immersion. 

54. Rear Aprons Decrease Uplift. Another case will now be 
examined. In Fig. 42 suppose the fore apron removed and a rear 
apron substituted. In this case the point of ingress of the percolat¬ 
ing water is thrown back from b to b' the hydraulic gradient is a c , 










































72 


DAMS AND WEIRS 


the triangle of hydrostatic uplift is b'a^c. This uplift from V to b 
is more than neutralized by the rectangle of water a'abb r , which 
overlies the rear apron; the latter is therefore not subject to any 
uplift and, owing to its location, is generally free from erosion by 
moving water, consequently it can be made of clay, which in this 
position is water-tight as concrete masonry. A glance at Fig. 42 
will demonstrate at once the great reduction of uplift against the 
base of the wall effected by the expedient of a rear apron, the uplift 
being reduced from a^bc to fbc, more than one-half. Thus a rear 



apron is a sure remedy for uplift while the fore apron, if solid, should 
be made as short as possible, or else should be formed of open work, 
as heavy slabs with open joints. In the rear of overfall dams stanch¬ 
ing clay is often deposited by natural process, thus forming an 
effective rear apron. Many works owe their security to this fact 
although it often passes unrecognized. 

55. Rock Below Gravel. Fig. 43 represents a dam founded 
on a stratum of pervious material beneath which is solid rock. A 
fore curtain wall is shown carried down to the impervious rock. 
The conditions now are worse than those resulting from the imper- 























DAMS AND WEIRS 


73 


vious fore apron in Fig. 42 as the hydraulic gradient and piezometric 
line are now horizontal. The reduced area of vertical hydrostatic 
pressure is 1066 against which the wall can only furnish 1200; there 
is, therefore, an effective area of only 134 to resist a water pressure 
at the rear of 800, consequently the wall must fail by sliding or 
overturning as the graphical stress lines clearly prove. The proper 
position of a diaphragm curtain wall is at the heel, not at the toe of 
the dam; in this location it will effectively prevent all uplift. In 
the case where an impervious stratum does not occur at a reasonable 
depth the remedy is to provide a long rear apron which will reduce 
hydrostatic uplift to as small a value as may be desired, or else a 
combination of a vertical diaphragm with a horizontal apron can be 



used. In many cases a portion only of the required rear apron need 
be provided artificially. With proper precautionary measures the 
deposit of the remaining length of unfinished apron can safely be 
left for the river to perform by silt deposit, if time can be afforded 
for the purpose. 

56. Gravity Dam Reinforced against Ice Pressure. This sec¬ 
tion will be concluded with a recent example of a gravity dam rein¬ 
forced against ice pressure, which is given in big. 44, viz, that of the 
St. Maurice River dam situated in the Rrovince of Quebec. The 
ice pressure is taken as 25 tons per foot run, acting at a level corre¬ 
sponding to the crest of the spillway, which latter is shown in Fig. 
58. The profile of Fig. 44 is pentagonal, the crest has been given 



























74 


DAMS AND WEIRS 


the abnormal width of 20 feet, while the base is f of the height, 
which is about the requirement, were ice pressure not considered. 
The horizontal ice pressure, in addition to that of the water upheld, 
will cause the line of pressure to fall well outside the middle third, 
thus producing tension in the masonry at the rear of the section. 
To obviate this, the back of the wall is reinforced with steel rods to 
the extent of 1| square inches per lineal foot of the dam. If the 
safe tensile strength of steel be taken at the usual figure of 16,000 



pounds, or 8 tons per square inch, the pull exerted by the reinforce- 
ment against overturning will be 12 tons per foot run. This force 
can be considered as equivalent to a load of like amount applied at 
the back of the wall, as shown in the figure. The section of the 
dam is divided into two parts at El 309 and the incidence of the 
resultant pressure at this level and at the base is graphically obtained. 
The line of pressure connecting these points is drawn on the pro¬ 
file. The line falls outside the middle third in the upper half of 
the section and within at the base, the inference being that the 

































DAMS AND WEIRS 


75 


section would be improved by conversion into a trapezoidal outline 
with a narrower crest and with some reinforcement introduced as 
has been done in the spillway section, shown in Fig. 58. 

It will be noticed that the reinforcement stops short at El 
275. This is allowed for by assuming the imposed load of 12 tons 
removed at the base of the load line in the force polygon. The 
line Rb starting from the intersection of i? 4 with a horizontal through 
El 275.0 is the final resultant at the base. This example is most 
instructive as illustrating the combination of reinforcement with a 
gravity section in caring for ice pressure, thus obviating the undue 
enlargement of the profile. 

GRAVITY OVERFALL DAMS OR WEIRS 

57. Characteristics of Overfalls. When water overflows the 
crest of a dam it is termed an overfall dam or weir, and some modi¬ 
fication in the design of the section generally becomes necessary. 
Not only that, but the kinetic effect of the falling water has to be 
provided for by the construction of an apron or floor which in many 
cases forms by far the most important part of the general design. 
This is so pronounced in the case of dwarf diversion weirs over wide 
sandy river beds, that the weir itself forms but an insignificant part 
of the whole section. The treatment of submerged weirs with aprons, 
will be given elsewhere. At present the section of the weir wall 
alone will be dealt with. 

Typical Section. Fig. 45 is a typical section of a trapezoidal 
weir wall with water passing over the crest. The height of the crest 
as before, will be designated by H , that of reservoir level above 
crest by d, and that of river below by D . The total height of the 
upper still water level, will therefore, be H +d. 

The depth of water passing over the crest should be measured 
some distance upstream from the overfall just above where the 
break takes place; the actual depth over the crest is less by reason 
of the velocity of the overfall being always greater than that of 
approach. This assumes dead water, as in a reservoir, in the upper 
reach. On a river or canal, however, the water is in motion and has 
a velocity of approach, which increases the discharge. In order to 

V 2 

allow for this, the head ( h ) corresponding to this velocity, or — 



76 


DAMS AND WEIRS 


multiplied by 1.5 to allow for impact, or h = .0233 F 2 , should be 
added to the reservoir level. Thus supposing the mean velocity 
of the river in flood to be 10 feet per second 100X.0233 or 2.3 
feet would have to be added to the actual depth, the total being 
15 feet in Fig. 45. The triangle of water pressure will have its 
apex at the surface, and its base will, for the reasons given 
previously, be taken as the depth divided by the specific gravity 
of the material of the wall. The triangle of water pressure will 



Fig. 45. Typical Section of Trapezoidal Weir Wall 


be truncated at the crest of the overfall. The water pressure acting 
against the back of the wall will thus be represented by a trapezoid, 


not a triangle, whose base width is 


Il+d 

P 


and its top width at crest 



Its area therefore (back vertical) will be 



d\H 

p) 2 ' 


If the back is inclined the side of the trapezoid becomes II i. The 
general formula is therefore 


A = (II or Ih)X 


(II+2d) 

2 P 







































DAMS AND WEIRS 


77 


same 


Hi being the inclined length of the back of the wall. The vertical 
distance of its point of application above the base according to 

formula (5) page 19 is h = ~ and will be the 

whether the back is vertical or inclined. 

58. Approximate Base Width. With regard to the drop wall 
itself, owing to the overfall of water and possible impact of floating 
timber, ice, or other heavy bodies, a wide crest is a necessity. A 
further strengthening is effected by adopting the trapezoidal profile. 
The ordinary approximate rule for the base width of a trapezoidal 
weir wall will be either 

(H+d) 


b = 




(16) 


or 


b = 


(7/+.6d) 

Vo 


(16a) 


The correctness of either will depend on various considerations, 
such as the value of cl, the depth of the overfall, that of hi or velocity 
head and also of Z), the depth of the tail water; the inclination given 
to the back, and lastly, whether the weir wall is founded on a porous 
material and is consequently subject to loss of weight from uplift. 
Hence the above formulas may be considered as approximate only 
and the base width thus obtained subject to correction, which is 
easiest studied by the graphical process of drawing the resultant on 
to the base, ascertaining its position relative to the middle third 
boundary. 

59. Approximate Crest Width. With reference to crest width, 
it may be considered to vary from 


k = V H-\-d 

(17) 

= V // + Vd 

(18) 


the former gives a width sufficient for canal, or reservoir waste weir 
walls, but the latter is more suitable for river weirs, and is quite so 
when the weir wall is submerged or drowned. 

In many cases, however, the necessity of providing space for 
falling shutters or for cross traffic during times when the weir is not 






78 


DAMS AND WEIRS 


acting, renders obligatory the provision of an even wider crest. 
With a moderate width, a trapezoidal outline has to be adopted, in 
order to give the requisite stability to the section. This is formed 
by joining the edge of the crest to the toe of the base by a straight 


line, the base width of 


(H+d) 

Vp 


being adopted, as shown in Fig. 45. 


When the crest width exceeds the dimensions given in formula 
(18), the face should drop vertically till it meets the hypothenuse 
of the elementary profile, as is the case with the pentagonal profile 
of dams. An example of this is given in Fig. 52 of the Dhukwa 
weir. The tentative section thus outlined should be tested by 
graphical process and if necessary the base width altered to conform 
with the theory of the middle third. 

In Fig. 45 is given a diagram of a trapezoidal weir 60 feet high 
with d —15 feet. According to formula (17) the crest width should 

be V75 = 8.7 feet, and according to (18), 7.74+3.87 = 11.6. An aver¬ 


age of 10 feet has been adopted, which also equals 



The profile 


therefore, exactly corresponds with the elementary triangle canted 
forward and truncated at the overfall crest. 

60. Graphical Process. In graphical diagrams, as has already 
been explained, wherever possible half widths of pressure areas are 
taken off with the compasses to form load lines, thus avoiding the 
arithmetical process of measuring and calculating the areas of the 
several trapezoids or triangles, which is always liable to error. There 
are, however, in this case, three areas, one of which, that of the 
reverse water pressure, has an altitude of only half of the others. 
This difficulty is overcome by dividing its half width by 2. If one 
height is not an exact multiple, as this is of II, a fractional value 
given to the representing line in the polygon will often be found to 
obviate the necessity of having to revert to superficial measures. 
The application of the reverse pressure Pi here exhibited is similar 
to that shown in Fig. 16; it has to be combined with R, which latter 
is obtained by the usual process. This combination is effected in 
the force polygon by drawing a line Pi equal to the representative 
area, or half width of the back pressure, in a reverse direction to P. 
The closing line Pi is then the final resultant. On the profile itself 
the force line Pi is continued through its center of gravity till it 



DAMS AND WEIRS 


79 


intersects R, from which point Ri is drawn to the base. If this por¬ 
tion of the face of the weir is very flat, as is sometimes the case, Pi 
may be so deflected as to intersect R below the base altogether as 
is shown in Fig. 50. In such event, Pi is drawn upward instead of 
downward to intersect the base. The effect of Pi is to throw the 
resultant Pi farther inward but not to any great extent. It inn 
proves the angular direction of P, however. 

Reverse Pressure. A dam is usually, but not invariably, exempt 
from the effect of reverse pressure. This reverse water pressure is 
generally, as in this case, favorable to the stability of the weir, but 
there are cases when its action is either too slight to be of service or 
is even detrimental. This occurs when the face of the weir wall is 
much inclined, which points to the equiangular profile being most 
suitable. An example illustrative of the above remarks is given 
later in Fig. 50 of the Folsam dam. 

As the moments of the horizontal pressure of water on either 
side of the weir wall vary almost with the cubes of their height, it 
is evident that a comparatively low depth of tail water will have but 
small influence and may well be neglected. When a vertical back 
is adopted, the slope is all given to the face; by which the normal 
reverse water pressure is given a downward inclination that reduces 
its capacity for helping the wall. 

61. Pressures Affected by Varying Water Level. Calculations 
of the depths of water passing over the weir or rather the height of 
reservoir level above the weir crest, designated by d, and of the 
corresponding depth D in the tail channel, are often necessary for 
the purpose of ascertaining what height of water level upstream, 
or value of d, will produce the greatest effect on the weir wall. In 
low submerged or drowned weirs, the highest flood level has often 
the least effect, as at that time the difference of levels above and 
below the weir are reduced to a minimum. This is graphically 
shown in Fig. 46, which represents a section of the Narora dwarf 
weir wall, to which further reference will be made in section 124, 
Part II. In this profile two resultant pressures, R and R u are 
shown, of which R u due to much lower water level of the two stages 
under comparison, falls nearer the toe of the base. 

The Narora weir, the section of the weir wall of which is so 
insignificant, is built across the Ganges River in Upper India at the 


80 


DAMS AND WEIRS 


head of the Lower Ganges Canal, Fig. 93. The principal part of 
this work, which is founded on the river sand, consists not in the low 
weir wall, although that is § mile long, but in the apron or floor, 
which has to be of great width, in this case 200 feet. 

As will be seen in Fig. 46 the flood level of the Ganges is 16 feet 
above the floor level, while the afflux, or level of the head water, is 
two feet higher. The river discharges about 300,000 second-feet 
when in flood. When full flood occurs, the weir is completely 
drowned, but from the diagrams it will be seen that the stress on 
the wall is less when this occurs than when the head water is 



much lower. This result is due to the reverse pressure of the 
tail water. 

The rise of the river water produces, with regard to the stress 
induced on the weir, three principal situations or ‘"stages” which are 
enumerated below. 

(1) When the head water is at weir crest level; except in cases 
where a water cushion exists, natural or artificial, the tail channel 
is empty, and the conditions are those of a dam. 

(2) When the level of the tail water lies below weir crest level 
but above half the height of the weir wall. In this case the recip¬ 
rocal depth of the head water above crest is found by calculation 
































































DAMS AND WEIRS 


81 


(3) At highest flood level, the difference between the head 
and tail water is at a minimum. In an unsubmerged weir or over- 
fall dam the greatest stress is generally produced during stage (3). 
In a submerged weir the greatest stress is produced during stage (2). 

62. Moments of Pressure. The moments of the horizontal 
water pressure on either side of a wall are related to each other in 
proportion to the cubes of their respective depths. In cases where 
the wall is overflowed by the water, the triangle of pressure of the 
latter, as we have seen, is truncated at the weir crest. The moment 
(M) of this trapezoidal area of pressure will be the product of its 
area with li, or the product of the expressions in formula (I) and 
formula (5) as follows: 


or 


M= II 


(H+2d) H (H+Sd) 
2 p 3 (II+2d) 


Xiv 


H ^w 

M=~ (H+Sd) 

bp 



That of the opposing tail water will be M = ——, the difference of 

bp 

these two being the resultant moment. For example, in the case 
shown in Fig. 46, during stage (1) 11 = 10, D = 0, unbalanced moment 

— — 7 r~ = 166.6—. In stage (2) 77 = 10, d = 3.5, and D = 10. Then 
bp p 


7/1 7/1 

the unbalanced moment will be — [(100 X 20.5) — 1000] = 175—. 

bp P 

In stage (3) 77 = 10, D = 16; <7 = 8, and D— II = 6 feet. There will 
thus be two opposing trapezoids of pressure, and the difference in 
their moments will be 


w (100 X 34) w (100 X 28) _ ^ ^ w 

6p 6p p 

Thus stage (2) produces the greatest effect, the least being stage 
(3). In this expression (w) symbolizes, as before, the unit weight 
of water, per cubic foot or, ^ ton. 

In spite of this obvious fact, many weir wall sections have been 
designed under the erroneous supposition that the overturning 
moment is greatest when the upper water is at crest level and the tail 
channel empty, i.e., at a time when the difference of levels above and 








FLOOD 57ASF 


82 


DAMS AND WEIRS 



below the weir is at a 
maximum, or at full 
flood when the differ¬ 
ence is at a minimum. 

63. Method of 
Calculating Depth of 
Overfall. During the 
second stage of the 
« river the value of d , 

p 

£ the depth of the over- 
I fall, will have to be 

03 

^ calculated. To effect 
| this the discharge of 
§ the river must first be 
H estimated when the 

u .. 

« surface reaches the 
1 crest level of the weir, 

Oh 

=3 which is done by use 

^ of the formula, Q = 

■« .— 

^ Hcvrs, given in sec- 

.2 tion 35, page 47 of 

| “Hydraulics, Ameri- 

m can School of Corre¬ 
ia 

I spondence”, A being 
“ the area, equal to d 
| times length of weir 
5 (c) Kutter’s coeffi- 

^ cient, (r) the hydraulic 
.sp mean radius, and s , 
the surface grade or 
slope of the river. The 
discharge for the whole 
river should now be 
divided by the length 
of the weir crest, the 
quotient giving the 
unit discharge, or that 
per foot run of the weir. 












































































































DAMS AND WEIRS 


83 


The depth required to pass this discharge with a free overfall 
is found by use of Francis formula of 3.33# or a modification of it 
for wide crest weirs for which tables are most useful. See “Hydrau¬ 
lics”, section 24, p. 30. 

For example, supposing the river discharge with tail water 
up to crest level is 20 second-feet per foot run of the weir. Then 

3.33# = 20. Whence # = 6 and d=Vfi 2 = 3.3 feet. This ignores 
velocity of approach, a rough allowance for which would be to 
decrease d by (hi) the velocity head, or by .0155 F 2 . 

64. Illustrative Example. Fig. 47 illustrates an assumed case. 
Here the weir is 15 feet high, 3 stages are shown: 

(1) When head water is at crest level; 

(2) When tail water is 7J feet deep, and the reciprocal depth of the 

head water is assumed as 4 feet; and 

(3) With tail water at crest level and head water assumed 7 feet 

deep above crest. 

The three resultants have been worked out graphically. From 
their location on the base the greatest stress is due to R 2 , i.e., 
stage (2). 

The hydraulic gradients of all three stages have been shown with 
an assumed rear and fore apron on floor. In (1) more than half the 
weir body lies below the piezometric line, which here corresponds 
with the hydraulic gradient, while in (2) nearly the whole lies below 
this line and in (3) entirely so. 

Owing to this uplift it is well always to assume the s.g. of a weir 
wall under these conditions as reduced by immersion to a value of 
p — 1. In these cases the triangles of water pressure are shown with 

their bases made —, or zr~r, instead of Actually, however, the 

p —1 1.4 2.4 

resistance of the weir wall to overturning relative to its base at floor 
level is not impaired by flotation, but as weight in these cases is a 
desideratum, the weir wall should be designed as if this were the case. 
The rear apron is evidently subject to no uplift, but the fore apron is, 
and its resisting power, i.e., effective weight, is impaired by flotation. 
See section 52 and also the later sections on “ Submerged Weirs in 
Sand”, Part II. 

65. Examples of Existing Weirs. Some examples of existing 
weirs will now be given. Fig. 48 is a profile of the LaGrange over- 





DAMS AND WEIRS 


• 84 


fall dam at the head of the Modesto and Tuolumne canals, Fig. 49. 
No less than 13 feet depth of water passes over its crest, 2 feet being 


El. 315 



Fig. 48. Profile of LaGrange Overfall Dam at Head of Modesto and Tuolumne Canals 


^ CAt1AL HEAD added to allow for velocity of ap¬ 

proach. It is built on a curve of 
300 feet radius. The graphical 
analysis of the section shows that 
the resultants (R.E.) and (R.F.) 
drawn on the profile fall within 
the middle third. In this process 
the reverse pressure due to tail 
water has been neglected. Its 
effect will be small. 

It is a doubtful point whether 
the reverse pressure actually exer¬ 
cised is that due to the full depth 
of the tail water. The overflow 
causes a disturbance and probably more or less of a vacuum at 
the toe of the weir wall, besides which the velocity of impact causes 
a hollow to be formed which must reduce the reverse pressure. In 
some instances, as in the case of the Granite Reef dam, Fig. 55, the 



Fig. 49. Location Plan of LaGrange Weir 































































DAMS AND WEIRS 


85 


effective depth of the tail water is assumed as only equal to that of 
the film of overflow. This appears an exaggerated view. How¬ 
ever, in a tiigli overfall dam, the effect of the reverse is often so small 
that it can well be neglected altogether. In cases where the tail 
water rises to f or more of the height of the dam its effect begins to 
be considerable, and should be taken into account. 

66. Objections to “Ogee” Overfalls. Professional opinion 
seems now to be veering round in opposition to the “bucket” or 
curved base of the fore slope which is so pronounced a feature in 
American overfall dams. Its effects are undoubtedly mischievous, 
as the destructive velocity of the falling water instead of being 
reduced as would be the case if it fell direct into a cushion of water, 
is conserved bv the smooth curved surface of the bucket. In the 
lately constructed Bassano hollow dam (see Figs. 84 and 85, Part II), 
the action of the bucket is sought to be nullified by the subsequent 
addition of baffles composed of rectangular masses of concrete fixed 
on the curved slope. The following remarks in support of this 
view are excerpted from “The Principles of Irrigation Engineering” 
by Mr. F. Id. Newell, formerly Director of United States Reclamation 
Service. “Because of the difficulties involved by the standing wave 
or whirlpool at the lower toe of overflow dams, this type has been 
made in many cases to depart from the conventional curve and to 
drop the water more nearly vertically rather than to attempt to 
shoot it away from the dam in horizontal lines.” 

67. Folsam Weir. Fig. 50 is of the Folsam weir at the head 
of the canal of that name. It is remarkable for the great depth of 
flood water passing over the crest which is stated to be over 30 feet 
deep. The stress lines have been put on the profile with the object 
of proving that the reverse pressure of the water, although nearly 
40 feet deep has a very small effect. This is due to the flat inclina¬ 
tion given to the lower part of the weir, which has the effect of 
adding a great weight of water on the toe where it is least wanted 
and thus the salutary effect of the reverse pressure is more than 
neutralized. The section is not too heavy for requirements, but econ¬ 
omy would undoubtedly result if it were canted forward to a nearly 
equiangular profile, and this applies to all weirs having deep tail water, 
and to drowned weirs. It will be noted that a wide crest allows 
but very little consequent reduction in the base width in any case. 


86 


DAMS AND WEIRS 


The stress diagram in Figs. 50 and 50a are interesting as show¬ 
ing the method of combining the reverse pressures with the ordinary 
Haessler’s diagram of the direct water pressure. The profile is 
divided into three parts as well as the direct water pressure, whereas 
the reverse pressure which only extends for the two lower divisions 
is in two parts. The stress diagrams present no novel features till 
R 2 is reached. This force on the profile comes in contact with reverse 
force 1" before it reaches its objective 3 1 . The effect of the reverse 



pressure is to deflect the direction of the resultant in the direction 
of i? 3 , which latter, as shown in the force polygon, Fig. 50a, is the 
resultant of 1", set out from the point 6, and of R 2 . The new result¬ 
ant R 3 continues till it meets 3 1 . The resultant of R 2 and 3 1 is the 
reverse line drawn upward to meet the vertical force 3, parallel to 
its reciprocal in Fig. 50a, which is the dotted line joining the termi¬ 
nation of 3 1 , i.e., (a) with that of 1". 

Following the same method the resultant R 4 is next drawn 
downward to meet 2", which latter in the force polygon is set out 























































DAMS AND WEIRS 


87 




from the ternination of the vertical (3). The resultant of P 4 and 
2" is the final P 5 . This line is drawn upward on the profile inter¬ 
secting the base at B. If the reverse pressure were left out of con¬ 
sideration, the force i ?2 would continue on to its intersection with 
3 1 and thence the reverse recovery line drawn to meet (3) will be 
parallel to ba (not drawn) in the force polygon. This reverse line 
will intersect the line (3) in the profile almost at the same spot 
as before. 

The final line will be parallel to its reciprocal ca (not drawn 
in Fig. 50a) and will cut the base outside the intersection of P 5 . To 
prevent confusion these lines have not been drawn on; this proves 
that the effect of the reverse pressure is detrimental to the stability 
of the wall, except in the matter of the inclination of P 5 . If the 
profile were tilted forward this would not be so. If Pi the resultant 
water pressure at the rear of the wall be drawn through the profile 
to intersect the resultant of all the vertical forces, viz, 1+2+3+r 1 
+?; 2 , this point will be found to be the same as that obtained by 
producing the final P 5 backwards to meet Pi. 

Determination of Pi. To effect this, the position of Pi has to 
be found by the following procedure: The load line db, Fig. 50a, is 
continued to /, so as to include the forces 3, Vi, and v 2 . The rays 
oc, oj, and ol are drawn; thus a new force polygon dot is formed to 
which the funicular, Fig. 50b, is made reciprocal. This decides the 
position of IF, or of 1+2+3, viz, the center of pressure (R. E.) as 
also that of W+Vi+v 2 which latter are the reverse pressure loads. 
The location of Pi is found by means of another funicular polygon C 
derived from the force polygon oad, by drawing the rays oa, of, and 
oe; Pi is then drawn through the profile intersecting the vertical 
resultant last mentioned at +. The line AB is then coincident 
with P 5 on Fig. 50. The vertical line through + is not N, i.e., is 
not identical with the vertical in Fig. 50, for the reason that N is 
the resultant of all the vertical forces, whereas the vertical in ques¬ 
tion is the centroid of pressure of all the vertical force less w u the 
weight of water overlying the rear slope of the wall. The location 
of N is found by drawing a horizontal P through the intersection 
of Pi with a line drawn through the c.g. of the triangle of water 
pressure w, this will intersect the back continuation of BA at c. 
A vertical CD through this point will correspond with that marked 


88 


DAMS AND WEIRS 




N in Fig. 50. The profile, Fig. 51, is a reproduction of that shown 
Fig. 50 in order to illustrate the analytical method of calculation or 
that by moments. 

68. Analytical Method. The incidence of the resultant R is 
required to be as ascertained on two bases, one the final base and 
the other at a level 13 feet higher. The section of the wall as before, 
is divided into three parts: (1) of area 840 square feet, (2) of 1092, and 
(3) of 838 square feet. The position of the c.g. of (1) is found by for¬ 
mula (7) to be 15.15 feet distant from a the heel of the base and will be 



Fig. 51. Diagram of Folsam Weir Illustrating Analytical Method of Calculation 


15.65 feet from b. That of (2) is 32.3 feet distant from its heel b. Th% 
reduced area of the water overlying the back down to b is estimated 
at 26 square feet and by formula (6) to be .5 feet distant from b. 
Again the reduced area of the reverse water overlying the fore 

IOC 

slope Vi is 92 square feet and the distance of its c.g. from b is 55 —— 

3 

= 48.8 feet. The moments of all these vertical forces equated 
with that of their sum (N) about the point b will give the position 
of N relative to b. 









































































DAMS AND WEIRS 


89 


Thus 

(1) 840X15.65 = 13146 

(2) 1092X23.3 =25443 

fa) 26 X .5 = 13 

fa) 92x48.8 = 4490 


2050X.t = 43092 = Moment of N 

x = 21 feet, nearly 


To obtain the distance f between 


N and R, f = 


M P —M Vl 

N 


Now 


the reduced area of P = 1257 and the height of the c.g. of the 
trapezoid having its base at b, and its crest level with that of the 
wall is calculated by formula (6), to be 22.1 feet. Again the reduced 
area of the reverse water pressure triangle pi is 120 square feet, 
the height of its c.g. above base is 8 feet. Consequently: 


- _ (1257X22.1) —(120X8) 
3 ~ 2050 


26820 

2050 


=13 feet 


For the lower base, the statement of moments about c is as fol¬ 
lows, t -2 being 240, and its distance 65 feet by formula (6). 


(N) 2050 X (21+ .3) =43665 

(3) 838X32.3 =27067 

fa) 10X .15 = 2 

fa) 240X65 =15600 

Total 3138 X;r =86334 


x— 


86334 

3138 


= 27.4 feet 


Now/i = ——— v^ Pl+p --, /i being the distance between Ni and ifa 

The value of Pi, the trapezoid of water pressure down to the base 
c, is 1747 square feet and the height of its c.g. by formula (19) or 
( 5 ) is 27 feet, that of (pi+jh) is 285 square feet and its lever arm 

— =12J feet. Then 
3 3 


(174 7X27) — (285 Xl2^) 
3138 


47169-3514 _ 43655 
3138 3138 


13.9 ft. 


The positions of N and N i being obtained, the directions of 
R and Ri are lines drawn to the intersections of the two verticals 















90 


DAMS AND WEIRS 


N and N i with two lines drawn through the c.g.’s of the trapezoid of 
pressure reduced by the moment of the reverse pressure, if any, or by 
(P—p). This area will consist, as shown in the diagram, of a trape¬ 
zoid superposed on a rectangle; by using formula (5) section 1, the 
positions of the c.g. of the upper trapezoid is found to be 12.58 feet 
above the base at a , while that of the lower is at half the depth of 
the rectangle, then by taking moments of these areas about b, the 
height of the c.g. is found to be 23.6 feet above the base at b, while 
the height for the larger area [Pi— (pi+2^)] down to c is 27 feet. 

In the graphical diagram of Fig. 51a the same result would be 
obtained by reducing the direct pressure by the reverse pressure 
area. Thus in the force diagram the vertical load line would remain 
unchanged but the water-pressure load line would be shorter being 
P—p and Pi— (pi+pz), respectively. This would.clearly make no 
difference in the direction of the resultants R and Pi and would save 
the two calculations for the c.g.’s of P and Pi. 

This weir is provided with a crest shutter in one piece, 150 feet 
long, which is raised and lowered by hydraulic jacks chambered 
in the masonry of the crest so that they are covered up by the gate 
when it falls. This is an excellent arrangement and could be imi¬ 
tated with advantage. The shutter is 5 feet deep. The width 


at base of lamina 2 of this weir is 55 feet, or very nearly 


H+d 
Vp ’ 


formula (16). 

69. Dhukwa Weir. A very similar work is the Dhukwa weir 
in India, Fig. 52, which has been recently completed. 

This overfall dam is of pentagonal section. Owing to the width 
of the crest this is obviously the best outline. 

The stress resultant lines have been drawn on the profile, which 
prove the correctness of the base width adopted. The tail water 
does not rise up to half the height of the weir. Consequently the 


formula is applicable in stage 3. 

Vp 


The effect of the tail water 


is practically nil. According to this formula the base width would 
2 

be 63X-^- = 42 feet, which it almost exactly measures—a further 
o 


demonstration of the correctness of the formula. The crest width 
should be, according to formula (18), V50+Vl3 = ll feet. The 




DAMS AND WEIRS 


91 


width of 17 feet adopted is necessary for the space required to work 
the collapsible gates. These are of steel, are held in position by 
struts connected with triggers, and can be released in batches by 
chains worked from each end. The gates, 8 feet high, are only 10 
feet wide. This involves the raising and lowering of 400 gates, the 
weir crest being 4000 feet long. The arrangement adopted in the 
Folsam weir of hydraulic jacks operating long gates is far superior. 
An excellent feature in this design is the subway with occasional side 
chambers and lighted by openings, the outlook of which is under¬ 
neath the waterfall, and has the advantage of relieving any vacuum 
under the falling water. 


AFFLUX fh El.903 



The subway could be utilized for pressure pipes and for cross 
communication, and the system would be most useful in cases where 
the obstruction of the crest by piers is inadvisable. The weir is 
4000 feet long and passes 800,000 second-feet, with a depth of 13 
feet. The discharge is, therefore, 200 second-feet per foot run of 


weir, which is very high. 


The velocity of the film will be 


200 

13 


15.4 


feet per second. With a depth of 13 feet still water, the discharge 
will be by Francis’ formula, 156 second-feet per foot run. To produce 
a discharge of 200 feet per second, the velocity of approach must be 
about 10 feet per second. This will add 2.3 feet to the actual value 






















































92 


DAMS AND WEIRS 


of d, raising it from 13 to 15.3 feet which strictly should have been 
done in Fig. 52. 

70. Mariquina Weir. Another high weir of American design, 
Fig. 53, is the Mariquina weir in the Philippines. It has the ogee 
curve more accentuated than in the LaGrange weir, 4 he stress 
lines have been drawn in, neglecting the effect of the tail water 
which will be but detrimental. The section is deemed too heavy 
at the upper part and would also bear canting forward with advan¬ 
tage, but there are probably good reasons why an exceptionally solid 



crest was adopted. The ogee curve also is a matter on which 
opinion has already been expressed. 

71. Granite Reef Weir. The Granite Reef weir over the 
Salt River, in Arizona, Figs. 54 and 55, is a work subsidiary to the 
great Roosevelt dam of which mention was previously made. 

It is founded partly on rock and partly on boulders and sand 
overlying rock. The superstructure above the floor level is the 
same throughout, but the foundations on shallow rock are remark¬ 
able as being founded not on the rock itself, but on an interposed 
cushion of sand. (See Fig. 54.) Reinforced concrete piers, spaced 
20 feet apart, were built on the bedrock to a certain height, to clear 






























DAMS AND WEIRS 


93 


all inequalities; these were connected by thin reinforced concrete 
side walls; the series of boxes thus formed were then filled level 
with sand, and the dam built thereon. This work was completed 
in 1908. The portion of the profile below the floor is conjectural. 
This construction appears to be a bold and commendable novelty. 
Sand in a confined space is incompressible, and there is no reason 
why it should not be in like situations. A suggested improvement 
would be to abandon the piers and form the substructure of two 
long outer walls only, braced together with rods or old rails encased 
in concrete. Fig. 55 is the profile on a boulder bed with rock below. 

72. Hydraulic Condi= 
tions. The levels of the 
afflux flood of this river are 
obtainable so that the st resses 
can be worked out. In most 
cases these necessary statis¬ 
tics are wanting. The flood 
downstream has been given 
the same depth, 12 feet, as 
that of the film passing over 
the crest. This is clearly 
erroneous. The velocity of 
the film allowing for 5 feet 
per second approach, is quite 
12 feet per second, that in 
the river channel could not 
be much over 5 feet, conse- 



Fig. 54. Section of Granite Reef Weir 
Showing Sand Cushion Foundations 


12X12 

quently it would require a depth of —-—- = 28 feet. The dam 


would thus be quite submerged, which would greatly reduce the 
stress. As previously stated, the state of maximum stress would 
probably occur when about half the depth of flood passes over 
the crest. However, the graphical work to find the incidence 
of the resultant pressure on the base will be made dependent on the 
given downstream flood level. After the explanations already 
given, no special comment is called for except with regard to the 
reverse water pressure. Here the curved face of the dam is altered 
into 2 straight lines and the water pressure consists of two forces 



























94 


DAMS AND WEIRS 


having areas of 17 and 40, respectively, which act through their 
c.g.’s. Instead of combining each force separately with the result¬ 



ant ( R ) it is more convenient to find their resultant and combine 
that single force with (i?.) This resultant P i must pass through 
the intersection of its two components, thus if their force lines are 








































































DAMS AND WEIRS 


95 


run out backward till they intersect, a point in the direction of Pi 
is found. P i is then drawn parallel to its reciprocal in the force 
polygon which is also shown on a larger scale at the left of the pro¬ 
file. The final resultant is R\ which falls just within the middle 
third of the base. R 2 is the resultant supposing the water to be 
at crest level only. The water in the river is supposed to have 
mud in solution with its s.g. 1.4. The base length of the triangle 

ot water pressure will then be --———--=-= 18.66. 

p 2.4 


The other water-pressure areas are similarly treated. If the rear 
curtain reaches rock the dam should not be subject to uplift. It 
could, however, withstand sub-percolation, as the hearth of riprap 
and boulders will practically form a filter, the material of the river 
bed being too large to be disintegrated and carried up between the 
interstices of the book blocks. The effective length of travel would 
then be 107 feet; add vertical 52 feet, total 159 feet, H being 20 feet, 

~ works out to = 8 which ratio is a liberal allowance for a boulder 
22 20 

bed. The fore curtain is wisely provided with weep holes to release 
any hydrostatic pressure that might otherwise exist underneath the 
dam. The Granite Reef dam has a hearth, or fore apron of about 
80 feet in width. A good empirical rule for the least width for a 
solid or open work masonry fore apron is the following: 


L = 2H+d (20) 

in which II is the height of the permanent weir crest above floor, 
and d is the depth of flood over crest. In this case II = 20, d= 12; 
least width of floor should then be 40+12 = 52 feet. The Bassano 
dam is 40 feet high with 14 feet flood over crest, the width of hearth 
according to this formula should be 94 feet, its actual width is 80 
feet which is admittedly insufficient. With a low submerged weir, 

formula (34), Part II, viz, Z = 3Vc//, will apply. Beyond the hearth 
a talus of riprap will generally be required, for which no rule can 
well be laid down. 

73. Nira Weir. Fig. 56 is of the Nira weir, an Indian work. 
Considering the great depth of the flood waterdown stream, the pro¬ 
vision of so high a subsidiary weir is deemed unnecessary, a water 
cushion of 10 feet being ample, as floor is bed rock. The section of 






96 


DAMS AND WEIRS 


the weir wall itself, is considered to be somewhat deficient in base 
width. Roughly judging, the value of Ii-\-d, on which the base 
width is calculated, should include about 3 or 4 feet above crest 
level. This value of d, it is believed would about represent the height 
of head water, which would have the greatest effect on the weir. 
The exact value of d could only be estimated on a knowledge of the 
bed slope or surface grade of the tail channel. The above estimate 

would make (H-\-d)= 36 feet, and with p = 2|, —4- =24 feet. 

_ __ 'P 

The top width, 8.3, is just Vi/+Vd, in accordance with the 
rule given in formula (18). 

A section on these lines is shown dotted on the profile. The 
provision of an 8-foot top width for the subsidiary weir is quite 



Fig. 56. Section of Nira Weir in India Showing Use of Secondary Weir 


indefensible, while the base width is made nearly equal to the 
height, which is also excessive. For purposes of instruction in the 
principles of design, no medium is so good as the exhibition of plans 
of actual works combined with a critical view of their excellencies 
or defects. The former is obtainable from record plans in many 
technical works, but the latter is almost entirely wanting. Thus 
an inexperienced reader has no means of forming a just opinion and 
is liable to blindly follow designs which may be obsolete in form 
or otherwise open to objection. 

74. Castlewood Weir. The Castlewood weir, Fig. 57, is of 
remarkable construction, being composed of stonework set dry, 
enclosed in a casing of rubble masonry. It is doubtful if such a 
section is any less expensive than an ordinary gravity section, or 































DAMS AND WEIRS 


97 


much less than an arched buttress dam of type C. Shortly after 
construction, it showed signs of failure, which was stated to be due 
to faulty connections with banks of the river; but whatever the 
cause it had to be reinforced, which was effected by adding a solid 
bank of earth in the rear, as shown in the figure. This involved 
lengthening the outlet pipes. In the overfall portion the bank must 
have been protected with riprap to prevent scouring due to the 
velocity of the approach current. 

75. American Dams on Pervious Foundations. In the United 
States a very large number of bulkhead and overfall dams and regu¬ 
lating works, up to over 100 feet in height have been built on foun¬ 



dations other than rock, such as sand, boulders, and clay. Most 
of these, however, are of the hollow reinforced concrete, or scallop 
arch types, in which a greater spread for the base is practicable 
than would be the case with a solid gravity dam. Whenever a core 
wall is not run down to impervious rock, as was the case in the 
Granite Reef Overfall dam, Fig. 55, the matter of sub-percolation 
and uplift require consideration, as is set forth in the sections on 
“Gravity Dams” and “Submerged Weirs on Sand”. If a dam 50 
feet high is on sand or sand and boulders, of a quality demanding 
a high percolation factor of say 10 or 12, it is clear that a very long 
rear apron and deep rear piling will be necessary for safety. 

All rivers bring down silt in suspension. When the overfall 
dam is a high one with a crest more than 15 or 20 feet above river- 










































I 


98 


DAMS i\ND WEIRS 


bed level, the deposit that is bound to take place in rear of the 
obstruction will not be liable to be washed out by the current, and 
additional light stanching silt will be deposited in the deep pool of 
comparatively still water that must exist at the rear of every high 
dam. For a low weir however this does not follow, and i? deposit is 
made it will be of the heavier, coarser sand which is not impermeable. 

The difficulty and expense of a long rear apron can be sur¬ 
mounted by the simple expedient of constructing only a portion of 
it of artificial clay, leaving the rest to be deposited by the river 
itself. To ensure safety the dam should be constructed and reser¬ 
voir filled, in two or three stages, with intervals between of sufficient 
length to allow the natural deposit to take place. Thus only a frac¬ 
tion of the protective apron need be actually constructed. Many 
works are in existence which owe their safety entirely to the fortu¬ 
nate but unrecognized circumstance of natural deposit having 
stanched the river bed in their rear, and many failures that have 
taken place can only be accounted for from want of provision for 
the safety of the work against underneath scour or piping and also 
uplift. The author himself once had occasion to report on the fail¬ 
ure of a head irrigation work which was designed as if on rock, 
whereas it was on a pervious foundation of boulders. When it 
failed the designers had no idea of the real cause, but put it down 
to a “treacherous river”, “ice move”, anything but the real reason, 
of which they were quite ignorant. Had a rear apron of sufficient 
width been constructed, the work would be standing to this day. 

76. Base of Dam and Fore Apron. The fore apron and base 
of an overfall dam or weir must be of one level throughout its 
length, if the foundation is of any other material than rock. The 
foundation core walls may have to vary more or less with the surface 
of the river bed, which is deep in some places, and shallow in 
others, but the apron level should be kept at or about low water 
level throughout. When a horizontal wall as an overfall dam is 
built across a river bed it obliterates the depressions and channels 
in it, the discharge over the weir is the same at all points or nearly 
so, consequently the tendency will be to level the bed downstream by 
filling the hollows and denuding the higher parts. 

Under these conditions it is evidently sheer folly to step up the 
apron to coincide with the section of the river bed, as the higher 


DAMS AND WEIRS 


99 


parts of the bed are bound to be in time washed out by the falling 
water and deposited in the deeper channels, and portions of the 
dam may easily be undermined. This actually occured in one case. 

77. Section of Spillway of St. Maurice River Dam. Fig. 58 is 
a section of the spillway portion of the reinforced bulkhead gravity 
dam, illustrated in Fig. 44. Owing to the absence of the heavy 
crest of Fig. 44, the back of the spillway profile is provided with 



Fig. 58. Diagram Showing Profile of Spillway Portion of Saint Maurice River Dam 

(See Fig. 44) 

double the amount of reinforcement shown in the former example. 
One half, viz, 1J inches, extends right down to the base, while the 
other half stops short at El 280. This is arranged for in the stress 
diagrams in the same way as explained in section 55, R$ being the 
final resultant on the base. The line of pressure falls slightly out¬ 
side the middle third in the upper half of the section. The effect 
would be to increase the tension in the reinforcement somewhat 
above the limit of 8 tons per square inch. The adoption of a trape¬ 
zoidal profile, would, it is deemed, be an improvement in this case 
as well as in the former. 


































OKHLA WEIR ON THE JUMNA RIVER, INDIA 

The view is taken from the sand bar downstream 











DAMS AND WEIRS 


PART II 


ARCHED DAMS 

78. General Characteristics. In this type, the whole dam, 
being arched in plan, is supposed to be in the statical condition of 
an arch under pressure. As, however, the base is immovably fixed 
to the foundations by the frictional resistance due to the weight of 
the structure, the lowest portion of the dam cannot possess full 
freedom of motion nor elasticity, and consequently must act more or 
less as a gravity dam subject to oblique pressure. 

However this may be, experience has conclusively proved that 
if the profile be designed on the supposition that the whole is an 
elastic arch, this conflict of stresses near the base can be neglected 
by the practical man. The probability is that both actions take 
place, true arch action at the crest, gradually merging into transverse 
stress near the base; the result being that the safety of the dam is 
enhanced by the combination of tangential and vertical stresses on 
two planes. 

In this type of structure, the weight of the arch itself is conveyed 
to the base, producing stress on a horizontal plane, while the water 
pressure normal to the extrados, radial in direction, is transmitted 
through the arch rings to the abutments. The pressure is, therefore, 
distributed along the whole line of contact of the dam with the sides 
as well as the ground. In a gravity dam, on the other hand, the whole 
pressure is concentrated on the horizontal base. 

Arch Stress. The average unit stress developed by the water 
pressure is expressed by the formula 

si = R “Short” Formula ( 21 ) 


• • 


b = 


RHw 

Si 


“Short” Formula (21a) 





102 


DAMS AND WEIRS 


in which R is the radius of the extrados, sometimes measured to the 
center of the crest, II the depth of the lamina, b its width, and w 
the unit weight of water or ^ ton. Into this formula p, the specific 
gravity of the material in the arch, does not enter. This simple 
formula answers well for all arched dams of moderate base width. 
When, however, the base width is considerable, as, sav, in the case 
of the Pathfinder dam, the use of a longer formula giving the maxi¬ 
mum stress ( 5 ) is to be preferred. This formula is derived from the 
same principle affecting the relations of s and s 1 , or of the maximum 
and average stresses already referred to in Part I on ‘‘Gravity Dams”. 
The expression is as follows, r being the radius of the intrados: 

2 R RHw 2 R 

,S‘ = ,S*i——-- =---X-7TT- 


“Long” Formula ( 22 ) 


“Long” Formula (22a) 

79. Theoretical and Practical Profiles. In a manner similar 
to gravity dams, the theoretical profile suitable for an arched dam 
is a triangle having its apex at the extreme water level, its base 
width being dependent on the prescribed limiting pressure. Success¬ 
ful examples have proved that a very high value for s, the maximum 
stress, can be adopted with safety. If it were not for this, the profit¬ 
able use of arched dams would be restricted within the narrow limits 
of a short admissible radius, as with a low limit pressure the section 
would equal that of a gravity dam. 

The practical profile is a trapezoid, a narrow crest being neces¬ 
sary. The water pressure area acting on an arched dam, is naturally 
similar to that in a gravity dam, the difference being, however, 
that there is no overturning moment when reverse pressure occurs 
as in a weir. The difference or unbalanced pressure acting at any 
point is simply the difference of the direct and the reverse forces. 
The areas of pressure on both sides, therefore, vary with the squares 
of their respective depths. 


or in terms of R and b 


s = 


(R+r) 


2 Hw 


L( 2 -L 

R\ R. 


also 










DAMS AND WEIRS 


103 


The water pressure on an arch acts normally to the surface of 
its back and is radial in direction; consequently the true line of pres¬ 
sure in the arch ring corresponds with the curvature of the arch and 
has no tendency to depart from this condition. There is, therefore, 
no such tendency to rupture as is the case in a horizontal circular 
arch subjected to vertical rather than radial pressure. This prop¬ 
erty conduces largely to the stability of an arch under liquid 
pressure. This condition is not strictly applicable in its entirety 
to the case of a segment of a circle held rigidly between abut¬ 
ments as the arch is then partly in the position of a beam. The 
complication of stress involved is, however, too abstruse for practical 
consideration. 

80 . Correct Profile. As we have already seen, the correct 
profile of the arched dam is a triangle modified into a trapezoid 
with a narrow crest. With regard to arch stresses, the most favorable 
outline is that with the back of the extrados vertical. The reason for 
this is that the vertical stress due to the weight of the arch, although 
it acts on a different plane from the tangential stresses in the arch 
ring, still has a definable influence on the maximum induced stress 
in the arch ring. The vertical pressure produces a transverse expan¬ 
sion which may be expressed as W xEXm, in which E is the coeffi¬ 
cient of elasticity of the material and m that of transverse dilation. 
This tends, when the extrados is vertical, to diminish the maximum 
stress in the section; whereas when the intrados is vertical and the 
back inclined, the modification of the distribution of pressure is 
unfavorable, the maximum stress being augmented. When the 
trapezoidal profile is equiangular, an intermediate or neutral condi¬ 
tion exists. A profile with vertical extrados should, therefore, be 
adopted whenever practicable. 

In very high dams, however, the pressure on the horizontal 
plane of the base due to the weight of the structure, becomes so 
great as to even exceed that in the arch ring; consequently it is 
necessary to adopt an equiangular profile in order to bring the center 
of pressure at, or near to, the center of the base, so as to reduce the 
ratio of maximum pressure to average pressure to a minimum. 
As stated in the previous section, when a vertical through the 
center of gravity of the profile passes through the center of the base, 
the maximum pressure equals the average, or s = s i. 


104 


DAMS AND WEIRS 


81. Support of Vertical Water Loads in Arched Dams. 

When the back of an arched dam is inclined, the weight of the 
water over it is supported by the base, the horizontal pressure of 
the water alone acting on the arch and being conveyed to the 
abutments. In the case of inclined arch buttress dams, however, a 
portion of the vertical load is carried by the arch, increasing its 
thrust above what is due to the horizontal water pressure alone. 
This is due to overhang, i. e., when the c. g. falls outside the base. 

82. Crest Width. The crest width of arched dams can be 
safely made much less than that of gravity dams and a rule of 


\ 2 ^ Rad. 335 



k=^U ( 23 ) 

would seem to answer the purpose, unless rein¬ 
forcement is used, when it can be made less. 


EXAMPLES OF ARCHED DAMS 

The following actual examples of arched 
dams will now be given. 

83. Bear Valley Dam. This small work, 
Fig. 59, is the most remarkable arched dam in 
existence and forms a valuable example of the 
enormous theoretical stresses which this type 
of vertical arch can stand. The mean radius 
being 335 feet according to formula (21) the 
unit stress will be 


R Hw 


= 60 tons, nearly 


Fig. 59. Section of Old Bear This section would be better if reversed. The 

actual stress is probably half this amount. 
This work has now been superseded by a new dam built below it, 
Fig. 77, section 103. 


84. Pathfinder Dam. This immense work, Fig. 60, is built 
to a radius of 150 feet measured to the center of the crest. That, 
however, at the extrados of the base of the section is 186 feet and this 
quantity has to be used for the value of R in the long formula (22). 
The unit stress then works out to 18 tons, nearly. The actual stress 
in the lowest arch ring is undoubtedly much less, for the reason 

















DAMS AND WEIRS 


105 



that the base must absorb so large a proportion of the thrust that 
very little is transmitted to the sides of the canyon. The exact 
determination of the proportion transmitted in the higher rings 
is an indeterminate problem, and the only safe method is to assume 
with regard to tangential arch stress that the arch stands cle^r of 

RAD. / 50 ' 




FAD. 186' ^ 


Fig. 60. Section of Pathfinder Dam 


the base. This will leave a large but indeterminate factor of safety 
and enable the adoption of a high value for s, the maximum unit 
stress. 

The profile of the dam is nearly equiangular in outline. This 
is necessary in so high a dam in order to bring the vertical resultants 
(W ) R. E. and ( N ) R. F. as near the center as possible with the 
object of bringing the ratio of maximum to mean stress as low as 
possible. 

The estimation of the exact positions of W and of N is made 
analytically as below. 













































106 


DAMS AND WEIRS 


There are only two areas to be considered, that of the water 
overlying the inclined back (v) and that of the dam itself (IF). 
Dividing v by 2\ (the assumed specific gravity of the material), 
reduces it to an equivalent area of concrete or masonry. 

210X31.5 


v 


2X2.25 
104 


= 1470 = 103 tons 


W = ^X210 = 10920 = 768 tons 

j-j 

Total, or N= 12390 = 871 tons 

Using formula (7), Part I, the c.g. of IP is 50.8 distant from the 
toe of the profile, then q or the distance of the incidence of IF from 

94 

the center point of the base is 50.8—— = 3.8. 

W 768 

The value of s l} or the mean unit stress is —, or —— =8.1 tons 

b 94 

6X3.8 1 0 . , T mW 0 . w0 , 

= 1.24; then 5 = —r— = 1.24X8.1 =10.1 


94 


and m = 1+^ = 1 
b 

tons. 

For Reservoir Full, to find the position of N, moments will be 
taken about the toe as follows 

Moment of 0 = 103x83.5= 8600 
Moment of IF = 768X50.8 = 39014 


Total N = 871 


= 47614 


. 47614 ... , 94 . , N 871 

then x = - =54.6; whence q = 54.6 — — = / . 6 feet and Si = ~ = 

= 9.26. By formula (9), Part I, m = l+^^^ = 1.48. 

94 

5 = 9.26X 1.48 = 13.7 tons 

From this it is evident that the unit stress in the base, due to 
vertical load only, is a high figure. It could be reduced by still 
further inclining the back; on the contrary, if the back were vertical 
N would equal W. Let this latter case be considered. The distance 
of the c. g. of the profile from the heel will then be by formula 
(7a), Part I 


x= T lb 




= 31.66 


94 


and the value of q will be — — 31.66 = 15.33 feet 















DAMS AND WEIRS 


107 


Then 


Si as before = — = 8.1 tons 
b 


02 

to = 1+£t = 1.98 and 
94 


s = 8.1 X 1.98 = 16 tons 


This stress is greater than that of N in the previous working which 
proves that the forward tilt given to these high dams is necessary 
to reduce the maximum unit stress on the base to a reasonable limit. 
A more equiangular profile would give even better results. 

85. Shoshone Dam. The Shoshone dam, Fig. 61, is designed 
on lines identical with the last example. It has the distinction of 
being the highest dam in the world but has recently lost this 



preeminence, as the Arrow Rock, quite lately constructed, Fig. 37, 
Part I, is actually 35 feet higher. This work is also in the United 
States. The incidents of the resultants Reservoir Empty and 
Reservoir Full, which will be explained later, have been shown 
graphically, and the analytical computation is given below. The 
vertical forces taken from left to right are (1), area 6480; (2), 
14,450; (3), water overlying back, reduced area 1880; total 22,810. 

Taking moments about the toe of the base, the distance of 
(1) is 54 feet, of (2) calculated by formula (7), Part I, is 58.3, and 
of (3) is 95 feet, roughly. 































108 


DAMS AND WEIRS 


Then (64S0 X 54) + (14450 X 58.3) + (1880 X 95) = 22810X». 
£ = 60 feet, nearly'. 


The value of q for N then is 60 


108 


2 


= 6 feet. 


Now = = =211, and bv formula (9), Part I, s = 211X 

b 108 

(108+36) OQ1 , + 281X2.4 Q1 , , 

-————=281 square feet =-—— = 21 tons, nearly. 

108 1 32 

The maximum arch unit stress by formula (22) is as follows: the 
' radius of the extrados of the base being 197 feet the fraction — = —— 


= .55 and II = 245 therefore s = 



2X245X1 

.55X1.45X32 


——• = 19.2 tons. 

2o.o 

Below the level 60 ft. above base, the stress on the arch does 
not increase. The arch stress is less than that due to vertical 
pressure N. This base should undoubtedly have been widened, 
the battered faces being carried down to the base, not cut off by 
vertical lines at the 60-foot level. 

Center of Pressure—New Graphical Method. In order to find the 
center of pressure in a case like Fig. 61, where the lines of forces (1) 
and (2) are close together, the ordinary method of using a force and 
funicular polygon involves crowding of the lines so that accuracy is 
difficult to attain. Another method now will be explained which is 
on the same principle as that of the intersection of cross lines used 
for finding the c. g. of a trapezoid. 

In Fig. 61, first the c.g.’s of the three forces are found (1) the 
water pressure area divided by p or 2.4 which equals 1880 square 
feet, (2) the upper trapezoidal part of the dam area 14,450, and (3) 
the lower rectangular area 6480. Then (1) is joined to (2) and this 
line projected on one side in any location as at b in Fig. Ola. 

From a, ac is set off horizontally equal to (2) or 14,450 and 
from b, bd is drawn equal to (1) or 1880; cd is then drawn and its 
intersection with ab at e gives the position of the resultant 1-2, which 
can now be projected on the profile at G. To obtain the resultant 
of the components (1-2) with (3) the line G -3 is drawn on the profile 










DAMS AND WEIRS 


109 


and a parallel to it drawn from e on Fig. Ola, intersecting the hori¬ 
zontal through (3) at /. From e, eg is laid off horizontally equal to 
(3) or 6480 and from/,/A equal to (1+2) or 1880+14,450 = 16,330. 
hg is then drawn and its intersection with ef at j is the centroid of the 
three forces, which projected on the profile to Gff on the line G -3 gives 
the location of the vertical resultant of 1+2 +3. 

86. Sweetwater Dam. The profile of the Sweetwater dam 
in California is given in Fig. 62. The original crest of the dam 



Dam, California 


was at El. 220 , or 95 feet above the base. Under these conditions 
the dam depended for its stability on its arched plan. If con¬ 
sidered as a gravity dam with allowable tension at the heel, the 
vertical pressure area is the triangle abe, here ^ = 16.5 and m works 

Tii AT" 

out to 3.15. A 7 = 226 tons and 5 = 46 feet whence s — —r— = 

b 


3.15X226 


46 


= 15.5 tons which is set down from a to b. 


2 N 

The tension at the heel = s 2 - 7 - = 15.5 — 9.82 = 5.7 tons 

b 













































110 


DAMS AND WEIRS 




Fig. 63. Location Plan of Sweetwater Dam, Showing Alterations 









































DAMS AND WEIRS 


111 


Value of S When Heel Is Unable to Take Tention. If the heel is 
unable to take tension, the pressure triangle will then be adc in which 
ac = 3 times the distance of the incidence of R from the toe, or 
3X6.5 = 19.5 feet and s is ob¬ 
tained by the following formula 

( 24 ) 


S = 


4 w N or W 
3 X b-2q 


here s = 


4X226 


= 23.2 tons 


3X46-33 

This dam has lately been raised 
to El 240, or by 20, feet and by 
the addition of a mass of concrete 
at the rear transformed into a 
gravity dam. The resultant due 
to this addition is Ri on the 
diagram, s works out to 10.6 
tons and there is no tension at 
the heel. Any bond between the 
new wall and the old has been 
studiously avoided. The new 
work is reinforced with cross bars and the rear mass tied into the 
superstructure. Fig. 63 is a plan of the dam as altered. 



Fig. 64. 


mmrnm 

Profile of Barossa Dam 



87. Barossa Dam. This dam, Fig. 64, is an Australian work, 
and although of quite moderate dimensions is a model of good 
and bold design. 


































112 


DAMS AND WEIRS 


The back is vertical and the fore batter is nearly 1 in 2.7. dhe 
outline is not trapezoidal but pentagonal, viz, a square crest imposed 
on a triangle, the face joined with the hypothenuse of the latter by 
a curve. The crest is slender, being only 4| feet wide, but is strength¬ 
ened by rows of 40-pound iron rails, fished together, built into the 
concrete. The maximum arch stress works out to 17J tons, the 
corresponding vertical stress on base to 6J tons. Fig. 65 is a site 
plan of the work. 

88. Lithgow Dam. Another example very similar to the last 
is the Lithgow dam, No. 2, Fig. 66. The arch stress in this works 
out by the short formula to nearly 13 tons; the radius is only 100 

feet, the vertical stress works out to 7 tons. 

Arched darns abut either on the solid 
rocky banks of a canyon or else on the end of 
a gravity darn. In cases where a narrow deep 
central channel occurs in a river, this por¬ 
tion can advantageously be closed by an 
arched dam, while the flanks on which the 
arch abuts can be gravity dams aligned 
tangential to the arch at each end. The dam 
will thus consist of a central arch with two 
inclined straight continuations. The plan of 
the Roosevelt dam, Fig. 26, Part I, will give 
an idea of this class of work. 

89. Burrin Juick Subsidiary Dam. In 
Fig. 67 is shown the profile of a temporary reinforced arched dam 
for domestic water supply at Barren Jack, or Burrin Juick, Australia. 
The reinforcement consists of iron rails. The unit arch pressure 
at the base works out to 21 tons, nearly. Reinforcement of perma¬ 
nent dams down to the base is not desirable, as the metal may cor¬ 
rode in time and cause failure, although the possibility is often 
stoutly denied. The main Burrin Juick dam is given in Part I, 
Fig. 36. 

90. Dams with Variable Radii. The use of dams of the type 
just described, is generally confined, as previously noted, to narrow 
gorges with steep sloping sides in which the length of the dam at the 
level of the bed of the canyon is but a small proportion of that at the 
crest. The radius of curvature is usually fixed with regard to the 



Fig. 66. Profile of Lithgow 
Dam 












DAMS AND WEIRS 


113 


length of chord at the latter level, consequently at the deepest 
level, the curvature will be so slight that arch action will be absent 
and the lower part of the dam will be subject to beam stresses, i.e., 
to tension as well as compression. In order to obviate this, in 
some recent examples the radius of curvature at the base is made 
less than that at the crest, and all the way up, the angle subtending 
the chord of the arc, which is variable 
in length retains the same measure 
throughout. This involves a change in 
the radius corresponding to the variable 
span of the arch. The further advantage 
is obtained, of reduction in the unit stress 
in the arch ring and in rendering the 
stress more uniform throughout. In 
very high dams, however, the base width 
cannot be much reduced as otherwise the 
limit stress due to the vertical loading 
will be exceeded. This arrangement of 
varying radii is somewhat similar to that 
used in the differential multiple arch 
given later. 

MULTIPLE ARCH OR 
HOLLOW ARCH 
BUTTRESS DAMS 

91. Multiple Arch Generally More 
Useful Than Single Arch Dams. It is 

evident that a dam which consists of a 
single vertical arch is suitable only for a 
narrow gorge with rock sides on which the 
arch can abut, as well as a rock bed; consequently its use is strictly 
limited to sites where such conditions are obtainable. A rock 
foundation is also essential for gravity dams, the unit compression 
on the base of which is too high for any material other than rock. 

The advantages inherent in the vertical arch, which are con¬ 
siderable, can however be retained by use of the so-termed multiple 
or scallop arched dam. This consists of a series of vertical or 
inclined arches, semicircular or segmental on plan, the thrust of 



Fig. 67. Profile of Burrin Juick 
Subsidiary Dam 





















114 


DAMS AND WEIRS 


which is carried by buttresses. The arrangement is, in fact, iden¬ 
tical with that of a masonry arched bridge. If the latter be con¬ 
sidered as turned over on its side, the piers will represent the but¬ 
tresses. In the case of a wide river crossing, with a bed of clay, 
boulders, or sand, the hollow buttressed and slab buttressed dams 
are the only ones that can well be employed with safety. The wide 
spread that can be given to the base of the structure in these two 
types enables the unit pressure on the base to be brought as low as 
from 2 to 4 tons per square foot. 

As has already been noticed in section 78, the arch is peculiarly 
well suited for economical construction. This is due to the fact 
that the liquid pressure to which the arch is subjected is normal 
to the surface and radial in direction. The pressure lines in the 
interior of the arch ring correspond with its curvature and con¬ 
sequently the arch can only be in compression; thus steel reinforce¬ 
ment is unnecessary except in a small degree near the crest in order 
to care for temperature stresses. In slab dams, on the other hand, 
the deck is composed of flat slabs which have to be heavily rein¬ 
forced. The spacing of the buttresses for slabs is limited to 15 
to 20 feet, whereas in hollow arch dams there is practically no 
limit to the spans which may be adopted. Another point is, that 
the extreme compressive fiber stress on the concrete in deck slabs 
is limited to five hundred to six hundred and fifty pounds per square 
inch; in an arch, on the other hand, the whole section is in com¬ 
pression which is thereby spread over a much greater area. For the 
reasons above given the arch type now under consideration should 
be a cheaper and more scientific construction than the slab type in 
spite of the higher cost of forms. 

92. Mir Alam Dam. The first example given is that of the 
Mir Alam tank dam, Fig. 68. This remarkable pioneer structure was 
built about the year 1806, by a French engineer in the service of 
H. H. the Nizam of Hyderabad in Southern India. The alignment 
of the dam is on a wide curve and it consists of a series of vertical 
semicircular arches of various spans which abut on short buttress 
piers, Fig. 69. The spans vary from 83 to 138 feet, the one in Fig. 
68 being of 122 feet. The maximum height is 33 feet. Water 
has been known to overtop the crest. The length of the dam is 
over 3000 feet. 





DAMS AND WEIRS 


115 


On account of the inequality of the spans, the adoption of the 
semicircular form of arch is evidently a most judicious measure, 
for the reason that an arch of this form under liquid pressure exerts 
no lateral thrust at the springing. The water pressure being radial 
in direction, cross pressure in the half arches in the line of the spring¬ 
ing is balanced and in equilibrium. Whatever thrust is exerted 
is not in the direction of the axis of the dam but that of the buttress 
piers. On the other hand, if the arches were segmental in outline 
the terminal thrust is intermediate between the two axes, and when 
resolved in two directions one component acts along the axis of 



Fig. 68. Plan of One Arch of Mir Alam Dam 
This remarkable pioneer dam was built in 1806, and consisted of 21 such arches. 


the dam. This has to be met, either by the abutment, if it is an 
end span, or else by the corresponding thrust of the adjoining half 
arch. The other component is carried by the buttress; therefore, 
if segmental arches are used, the spans should be equal in order to 
avoid inequality of thrust. Longer buttresses will also be requisite. 
The whole of this work is built of coursed rubble masonry in lime 
mortar; the unit stress in the arch ring at the base, using the short 

formula (21), wor ks out to 68 ] ^ 3 ^ — = 5 tons, nearly. The 

dam, therefore, forms an economical design. 













































116 


DAMS AND WEIRS 


The buttress piers are shown in section in Fig. 70, the section 
being taken through AB of Fig. 68. In this work the buttress 
piers are very short, projecting only 25 feet beyond the spring line 
of the arches, and being altogether only 35 feet long. This length 
and the corresponding weight would clearly be inadequate to with¬ 



stand the immense horizontal thrust which is equivalent to 


H 2 
2 


Iw = 


33 2 X1X146 


2X32 


= 2500 tons, nearly. 


It is evident that if the buttress pier slides or overturns, the 
arches behind it must follow, for which reason the twx> half arches and 
the buttress pier cannot be considered as separate entities but as 
actually forming one whole, and consequently the effective length 
of the base must extend from the toe of the buttress right back 
to the extrados of the two adjoining arches. At, or a little in the 



Fig. 70. Section of Buttress Pier of Mir Alam Dam Taken through AB of Fig. 68 


rear of the spring line, the base is split up into two forked curved 
continuations. The weight of these arms, i.e., of the adjoining 
half arches, has consequently to be included with that of the but¬ 
tress proper when the stability of the structure against overturning 
or sliding is estimated. 










































DAMS AND WEIRS 


117 


93. Stresses in Buttress. In the transverse section, Fig. 71, 
taken through CD of Fig. 68, the graphical calculations establish 
the fact that the resultant line R intersects the base, thus lengthened, 
at a point short of its center; the direction of the resultant R is 
also satisfactory as regards the angle of frictional resistance. 

Ri is the resultant on the supposition that the buttress is 
nonexistent. Its incidence on the base proves that the arch is 
stable without the buttress, which is therefore actually superfluous. 
With regard to sliding on the base, P = 2500 and W = 6828 tons. 



Fig. 71. Transverse Section of Mir Alam Dam Taken through CD, Fig. 68 


The coefficient of friction being .7 the factor of safety against sliding 
is nearly 2. If the arch were altered on plan from a semicircle 
to a segment of a circle, the radius would of necessity be increased, 
and the stress with it; a thicker arch would, therefore, be required. 
This would not quite compensate for the reduced length of arch, 
but on the other hand, owing to the crown being depressed, the 
effective base width would be reduced and would have to be made 
good by lengthening the buttress piers. W hat particular dis¬ 
position of arch and buttress would be the most economical is a 


































118 


DAMS AND WEIRS 




matter which could only be worked out by means of a number of 
trial designs. The ratio of versed sine to span should vary from 
j to Arcs subtending from 135 to 120 degrees are stated to be 
the most economical in material. 

94. Belubula Dam. There are not as yet very many modern 
examples of arch buttress dams, but each year increases their num- 








(<0 



Fig. 72. Profile Sections and Force Diagram for Belubula Dam, New South Wales 


ber. The Mir Alam dam has remained resting on its laurels without 
a rival for over 100 years, but the time has come when this type 
is being largely adopted. Fig. 72 shows an early example of a 
segmental panel arch dam. It is the Belubula dam in New South 
Wales. The arch crest is 37 feet above the base, very nearly the 
same as in the last example. The arches, which are inclined 60 degrees 
to the horizontal are built on a high solid platform which obliterates 






































































































DAMS AND WEIRS 


119 


inequalities in the rock foundation. This platform is 16 to 23 feet 
high, so that the total height of the dam is over 50 feet. The spans 
are 16 feet, with buttresses 12 feet wide at the spring line, tapering 
to a thickness of 5 feet at the toe; they are 40 feet long. The buttress 
piers, which form quadrants of a circle in elevation, diminish in 
thickness by steps from the base up, these insets corresponding 
with similar ones in the arch itself. These steps are not shown 
in the drawing; the arch also is drawn as if in one straight batter. 
The arches are elliptical in form, and the spandrels are filled up 
flush with the crown, presenting a flat surface toward the water. 

Some of the features of this design are open to objection: First, 
the filling in of the arch spandrels entirely abrogates the advantage 
accruing to arches under liquid pressure. The direction of the water 
pressure in this case is not radial but normal to the rear slope, thus 
exactly reproducing the statical condition of a horizontal arch 
bridge. The pressure, therefore, increases from the crown to the 
haunches and is parabolic, not circular, in curvature. The arches 
should have been circular, not elliptical, and the spandrels left 
empty to allow of a radial pressure which partly balances itself. 
Second, the stepping in of the intrados of the arch complicates the 
construction. A plain batter would be easier to build, particularly 
in concrete. Third, the tapering of the buttress piers toward the 
toe is quite indefensible; the stress does not decrease but with the 
center of pressure at the center of the base as in this case, the 
stress will be uniform throughout. 

95. Inclination of Arch to Vertical. The inclination of the axis 
of the arch to the vertical is generally a desirable, in fact, a necessary 
feature when segmental arched panels are used; the weight of water 
carried is of value in depressing the final resultant line to a suitable 
angle for resistance to shearing stress. As noted in section 90, 
the weight of the water overlying the arch does not increase the 
unit stress in the arch ring. Consequently, any inclination of 
axis can be adopted without in any way increasing the unit stresses 
due to the water pressure. 

When an arch is vertical it is clear that the water pressure is 
all conveyed to the abutments and the weight of the arch to its 
base. When an arch lies horizontally under water pressure both 
the weight of the water and that of the arch itself are conveyed 


120 


DAMS AND WEIRS 


to the abutment; when in an intermediate position part of the weight 
of the arch is carried to the base and part to the abutments. 

With regard to water pressure, the thrust being normal to 
the extrados of the arch the whole is carried by the abutments. 
In the case of arches which do not overreach their base the weight 
of water overlying the inclined back is conveyed to the base. In 

(RIIw) 


any case the unit stress in the arch 


cannot exceed that due 


to horizontal thrust. The total water pressure is greater with an 
inclined back, as the length of surface acted on is increased. In 
the diagram, Fig. 72a, the vertical load line W represents the 
weight of one unit or one cubic foot of the arch ring which is equal 
to wp. This force is resolved in two directions, one p, parallel 
to the axis of the arch, and the other n, normal to the former. The 
force n = W sin 6, d being the inclination of the arch axis to the 
vertical and p = W cos 6. The unit stress developed by the radial 
force n is similar to that produced by the water pressure which is 
also radial in direction and is 7?in; but 7?i, the radius in this case, 
is the mean radius, the pressure being internal, not external. The 
unit stress Si will then be 

Si = Riwp sin 6 ( 25 ) 

When d is 30° sin # = -!-; when 45°, sin 6 = —. 

2 3 


It will easily be understood that this unit stress due to n does 
not accumulate, but is the same at the first foot depth of the arch 
as it is at the bottom; the width of the lamina also does not affect 
it. However, the component p does accumulate, and the expression 
wp cos 6 should be multiplied by the inclined height H lf lying 
above the base under consideration. As H 1 =H sec 6, the unit 

compressive stress at the base will be f [ n which by is the mean 

b 


width of the arch. If the arch were a rectangle, not a trapezoid, 
s would equal IIwp simply. 

96. Ogden Dam. The Ogden dam, the profile and sectional 
details of which are shown in Fig. 73, is a notable example of the 
arch and buttress type. Its height is 100 feet. The inclination 
of the arches is less than \ to 1, or about 25 degrees to the vertical. 




DAMS AND WEIRS 


121 


The profile of the buttress is equinangular except for a small out- 
throw of the toe. On the whole it must be pronounced a good 
design, but could be improved in several particulars. For example, 
the arch is unnecessarily thick at the crest, and could well be reduced 
from 6 to 2 feet, thus effecting considerable economy. The designers 
were evidently afraid of the concrete in the arch leaking, and so 
overlaid the extrados with steel plates. A greater thickness of 
arch causing it to possess less liability to percolation under pressure, 
could have been provided by increasing the span and radius of the 



arches. The design consequently would be improved by adopting 
larger spans, say 100 feet; buttresses, say, 25 feet thick, their length 
being dependent on the width of base required to provide sufficient 
moment of resistance; and further, the inclination of the arches 
might require increasing to bring the center of pressure at, or close 
to the center of the buttress. The finish of the crest by another 
arch forming a roadway is an excellent arrangement, and is well 
suited for a bulkhead dam; for an overfall, on the other hand, the 
curved crest is preferable on account of the increased length of 
overflow provided. The stress diagram shows that the value of the 


























































































































122 


DAMS AND WEIRS 


vertical load N is 155,000 cubic feet or 10,598 tons, p being taken 
at 2J. The incidence of R on the base, is 5 feet from the center, 
whence q = 5, and by formula (9), Part I 

6 q\ 10598 


-£x 


( 1+ f)- 


110X16 110 


^ 8.91 tons 


the dimensions of A, the area of the base, being 110X16 feet. The 
pressure on the arch ring at the base by the short formula works 
24X100 


out to 


8X32 


= 9.4 tons. 


The contents of the dam per foot run amounts to 


104,500 

48 


2,177 cubic feet; that of a gravity dam would be about 3,500 cubic 
feet per foot run, making a saving in favor of the arched type of 
nearly 30 per cent. With a better disposition of the parts as indi¬ 
cated above, the saving would be increased to 40 or 50 per cent. 
Actually the saving amounted to only 12 per cent; this was owing 
to the steel covering which, as we have seen, could have been 
dispensed with. 

97. Design for Multiple Arch Dam. Fig. 74 is a design for 
a segmental arch panel dam, or rather, weir. The height of the 
crest is 64 feet above base with 5 feet of water passing over; the 
apex of the triangle of water pressure will then be 69 feet above 
the base. The inclination given the axis, which is coincident with 
that of the spring line and the intrados, is 60 degrees with the horizon. 

In designing such a work, the following salient points first 
require consideration. 

(1) Width of Span. This, it is deemed for economical reasons 
should be not less than the height of crest unless the state of the 
foundation requires a low unit stress. In the Mir Alam dam 
the span is over four times the depth of water upheld. In the present 
case it will be made the same, that is, 64 feet. 

(2) Thickness of Buttress Piers. As with bridge piers, the 
width should be at least sufficient to accommodate the skew-backs 
of the two arches; a width of 12 feet or about \ span will effect this. 

(3) Radius and Versed Sine. The radius will be made 40 
feet; this allows a versed sine of \ span, or 16 feet, which is con¬ 
sidered to be about the flattest proportion to afford a good curva- 









— 


H08V 





03 A// / 

HOMO' 









Fig. 74. Profile Plans and Sections of Segmental Arch Panel Dam 





























































































































124 


DAMS AND WEIRS 


ture, the greater the length of the arc, the more its condition will 
approximate to that of a circular arch, under liquid pressure. 

(4) Thickness of Arch. This must first be assumed, as its 
thickness depends on R, the radius of the extrados, as well as on 
the value assigned to s 1 , the limiting pressure. This latter will 
be fixed at below 15 tons, a value by no means excessive for arches 
under liquid pressure. With a base width of 7 feet, the radius 
of the extrados will be 47 feet. The base will be considered, not 
at the extreme depth of 64 feet below crest, but at the point marked 
D, where a line normal to the base of the inclined intrados cuts 
the extrados of the arch. II will, therefore, be 60 feet, allowing 
for the reverse pressure. The stress due to the water pressure, 
using the short formula ( 21 ), section 78, will be 


Si = 


R Hw 
b 


47X60X1 

7X32 


= 12.6 tons 


To this must be added that due to the weight of the arch ring from 
formula (25), Si = Riwp sin 6 (the angle 6 being 30° and its sine = J), 

43.5X3 

which in figures will be ■ ‘ ^ =1.6 tons, the total stress being 

2X40 6 


a trifle over 14 tons. The 7-foot base width will theft be adopted, 
p is taken as 2.4 and wp = ±\ ton. The depth of water producing 
this pressure is taken as 60, not as 65, feet which is ( II+d), the reason 
being that the reverse pressure due to the tail water, which must 
be at least level with the water cushion bar wall, will reduce the 
effective depth to 60 feet, during flood conditions. 

98. Reverse Water Pressure. The influence of the reverse 
pressure of water is much more considerable when hydrostatic 
pressure alone is exerted than is the case with overturning moment. 
In the case of an upright arch acting as an overfall weir the pres¬ 
sure of the tail water effects a reduction of the pressure to the extent 
of its area. Thus if A be the area of the upstream water pressure, 
and a that of the downstream, or tail water, the unbalanced pressure 
will be their difference, or A-a, and will vary as the square of their 
respective depths. When overturning moment is concerned, the 
areas have to be multiplied by a third of their depths to represent 
the moment on the base. The difference of the two will be in 
that case as the cubes of their respective depths. 





DAMS AND WEIRS 


125 


99. Crest Width of Arch. The crest width of the arch, accord¬ 
ing to formula (23), should be jV// = 3? feet, nearly. It will be 
made 3 feet, with a stiffening rib or rim of 3 feet in width. The 
crest width could be made proportional to the base width, say .3 b, 
and if this falls below 2 feet, reinforcement will be required. 

The length of the pier base is measured from the extrados of 
the arch, the two half arches forming, as already explained in section 
92, a forked continuation of the buttress pier base. 

The battering of the sides of the pier would clearly be a correct 
procedure, as the pressure diminishes from the base upward. A 
combined batter of 1 in 10 is adopted, which leaves a crest width 
of 5.6 feet. The length of the pier base, as also its outline, were 
determined by trial graphical processes, with the object of maneu¬ 
vering the center of pressure as near that of the base as pos¬ 
sible, so as to equalize the maximum and the mean unit stress as 
much as possible. This has been effected, as shown by the inci¬ 
dence of the final resultant on the elevation of the buttress pier. 

100. Pressure on Foundations. The total imposed weight 
is measured by N in the force diagram, and is equivalent to 150,000 
cubic feet of masonry, which at a specific gravity of 2.4 is equal to 


150,000X3 

40 


= 11,250 tons. 


The average pressure is this quantity 


divided by the area of the base, or by 125X12 = 1500 square feet, 
the quotient being 7\ tons, nearly. The maximum pressure will 
be the same owing to the incidence of R at the center of the base. 
This 7\ tons is a very moderate pressure for a hard foundation; 
if excessive, additional spread should be provided or else the spans 
reduced. It will be noticed that N greatly exceeds W. This is 
due to the added weight of water represented by the inclination 
given to the force line P, which represents the water pressure. 

Economy of Multiple Arches. The cubic contents per foot 

run work out to = 850 cubic feet, nearly, the denominator in 

/6 

the fraction being the distance apart of the centers of the buttress piers. 

2 

The contents of a gravity weir with base width - (H+d) and 

O 


top y/H+d, works out to 1,728 cubic feet; the saving in material 
is therefore over 50 per cent. 





126 


DAMS AND WEIRS 


101. Differential Arches. Fig. 75 is a study of a differential 
buttress arch weir. The principle of the differential arch consists 
in the radius increasing with the height of the arch, the unit 
stress is thus kept more uniform, and the stress area corre¬ 
sponds more closely with the trapezoidal profile that has necessarily 
to be adopted, than is the case when a uniform' radius is adopted as 
in Fig. 74. 

The arches are supposed to stand on a concrete or masonry 
platform ten feet high above the deepest part of the river bed, 
so that sluices if required could be provided below L.W.L. which 
is identical with the floor or fore apron level. The height is 35 feet 
to crest level. The depth of film passing over the crest is assumed 
at 5 feet and the reciprocal depth of tail water is 12 feet. Graphical 
analyses will be made at two stages, first, when water is at crest 
level and the river channel below is empty, second, at full flood. 
The inclination given to the intrados of the arch is 3 vertical to 2 
horizontal. The buttresses are placed 31 feet centers, allowing 
a span of 25 feet at base, here they are 6 feet wide, tapering to 2 
feet at crest. The span of the arch thus gradually widens from 
25 to 29 feet. The versed sine of the arc is made 5 feet at base 
and 2\ feet at crest. The radii at these positions are therefore 18.1 
and 43.3, respectively, measured to the intrados of the arch. These 
radii are horizontal, not normal to the intrados as in Fig. 73, and 
thus vary right through from 18.1 to 43.3 corresponding to the 
altered versed sine which decreases from 5 to 2\ feet, that half 
way up being 22 feet. 

The thickness of the arch at base is made 2 feet. 

Arch Unit Stress. Taking the base radius as 18.1, the base 

unit stress due to water pressure will be by formula (21), s = - • 

b ’ 


s = 


adding that due to the transmitted weight of the arch, formula (25), 
= Rw (f +p sin 0 , sin 6 being .6, the expression becomes 

18.lw £y+(2.4X.6) j , p being taken at 2.4, w at ^ ton, whence 

s = 10.4 tons, a moderate stress for a vertical arch. 

The real thickness of the arch is more like 2\ feet than 2 feet 
as properly it should be measured horizontally, not normally. 



DAMS AND WEIRS 


127 









































































































































128 


DAMS AND WEIRS 


Load Line. In the force polygon the load line is made up of 
five weights: (1) that of the overlying water has a content of 

X3o _ 1386Q cubic feet, equivalent to 433 tons; (2) the arch 

2 

145 tons; (3) the contents of the pier underlying the arch is found 
by taking the contents of the whole as if the sides were vertical 
and deducting the pyramid formed by the side batters. Thus the 

22 vfi y 35 

contents of the whole is ——-—— =2310, that of the pyramid is 


22X2X35 


= 513, difference 1797, or 135 tons; (4) the weight of 


the horizontal arch of the crest of the weir, 12 tons; (5) the contents 
of the buttress, by the prismoidal formula comes to l (Ai+ 4A m -j- 
Az) in which A i and A 2 are areas of the ends and A m of the middle 
section. Here A i = 0, A m = —~ X— = 52.5, and A 2 = 35X4 = 140; 

therefore (5) = —X35X[0+ (4X52.5) + 140] = 2042 cubic feet equiv- 

6 

alent to 153 tons. 

The total load foots up to 878 tons. 

H 2 l (35) 2 

P the horizontal water pressure = w— X / = — X " - X31 =593 

Z oZ z 

tons. 


The position of the several vertical forces is obtained as follows: 
That of 1, a triangular curved prism is at J its horizontal width; 
of 2 is found by formula (7), Part I, and by projection of this level 
on to the plan. The position of 3 has to be calculated by moments 
as below. 

The lever arm of the whole mass including the battered sides 
is at J width from the vertical end of 7J feet while that of the pyram¬ 
idal batter is at J the same distance, or 5| feet. 

The statement is then 


2310X7J = (1797Xx) + (514X5.5) whence £ = 7.84 feet 
The position of 5, the battered sloping buttress is obtained by 
taking the center part 2 feet wide and the outer side batters sep- 

1 35 

arately. The c.g. of the former is at — the length, —- = 11§ from the 

o o 








DAMS AND WEIRS 


129 


vertical end, and its contents are —pX2 = 1225 cubic feet = 92 tons. 

£ 

The weight of the whole is 153 tons, so that the side batters will 

q r 

weigh 153 — 92 = 61 tons, and be -y = 8.75 feet distant from the end. 

4 

Taking moments about the vertical end, we have 

153.r = (92 X11.67) + (61X 8.75) 


1606 in _, 
x = ——- = 10.5 feet 
153 


Therefore, the incidence of the resultant on the base line meas¬ 
ured 6.5 feet upstream from the center point. 

In Fig. 75a, A 7 = 878 tons and — = 14; b being 63 feet and 

q — 6.5 feet, whence m = 1.62 and the stress on the buttress, if only 1 

foot wide, =14X1.62 = 22.7 tons. The compression at the toe 
2N 

=—i - s= (28 — 22.7) =5.3 tons. These quantites have now to be 


divided by the base widths to obtain the unit stresses, which are as 

22.7 14 5 3 

follows: at heel, ^— = 3.8; at center, — = 2.3; at toe, —^~= 2.6 tons. 

This stress area is shown hatched in Fig. 75f. 

This stress diagram is useful as showing that owing to the 
incidence of P being behind the center point the total stress dimin¬ 
ishes toward the toe of the buttress, consequently it should be 
tapered on plan, as has been done. In Fig. 74 it has been shown 
that the stress being uniform by reason of the incidence of R at 
the center point of the base, the buttress has been made rectangular 
in plan at its base. The indicated unit stresses are very light 
which is a great advantage on a bad foundation. 

102. Flood Pressures. The second, or flood stage, will now be 
investigated. Here the vertical load line N in Fig. 75e is increased 
by 140 tons, the additional weight of water carried by the arch. The 
horizontal water pressure Pi is now 763 tons and A r =1018, their 
resultant being Pi. The reverse pressure due to a depth of 
12 feet of water is 70 tons, this combined with Pi, in Fig. 75a, 
results in P 2 the final resultant. The value of 6 is 35° 15' which is 
satisfactory. As q scales 5 feet, the unit stresses work out as 
follows: 











9tU9;iZ\ 


Fig. 76. Plan and Sectional Elevation of Big Bear Valley Dam 























































































































DAMS AND WEIRS 


131 


At heel 3.9 tons, at center 2.7, and at toe, 4.2 tons. 

The stress in the arch under a head of 38 feet comes to 11.5 
tons. Thus the stresses in stage 2 are higher than is the case with 
stage 1. 

At the end of a series of these scallop arches near either abut¬ 
ment the thrust of the arch resolved axially with the weir has to 
be met either by tying the last two arches by a cross wall and rein¬ 
forcing rods, or abutting the arch on an abutment supported by 
wall or a length of solid dam. This design would, it is considered, 
be improved if the versed sine of the arcs were made somewhat 
greater, as the arches are too flat near the crest. 

The following remarks bear on the curvature of the arch men¬ 
tioned in section 101. When a segmental arch is inclined, the 
spring line is at a lower level than the crown, consequently the 
water pressure is also greater at that level. But the thickness 
should vary with the pressure which it does not in this case. This 
proves the advisability of making the circular curvature horizontal, 
then a section at right angles to the inclined spring line will be an 
ellipse, while a horizontal section will be a segment of a circle. 
The reverse occurs with arches built in the ordinary way. There 
appears to be no practical difficulty in constructing forms for an 
inclined arch on this principle. 

103. Big Bear Valley Dam. Fig. 76 is a plan and sectional 
elevation of the new Bear Valley reinforced concrete multiple 
arch dam which takes the place of the old single arch dam men¬ 
tioned in section 83. The following description is taken from 
“Engineering News”, from which Fig 78 is also obtained. 

The new dam consists of ten arches of 30| feet, clear span at 
top, abutting on eleven buttresses. The total length of the dam 
is 363 feet on the crest; its maximum height from crest to base is 
92 feet (in a pocket at the middle buttress only), although, as the 
elevation in Fig. 76 shows, the average height of the buttresses 
is much less than that figure. The water face of the structure 
and the rear edge of the buttresses are given such slopes as to bring 
the resultant of the water-pressure load and that of the structure 
through the center of the base of the buttresses at the highest 
portions of the dam, Fig. 79. The slope for the water face 
up to within 14 feet of the top is 36° 52' from the vertical, and 


132 


DAMS AND WEIRS 


from that point to the crest is vertical. The slope of the down¬ 
stream edges of the buttresses is 2 on 1 from the bottom to the top, 
the vertical top of the face arches giving the piers a top width of 10 
feet from the spring line to the back edge. r lhe buttresses are 1.5 
feet thick at the top and increase in thickness with a batter of 
0.016 feet per foot of height or 1 in 60 on each side to the base for 
all heights. The arch rings are 12 inches thick at the top and 
down to the bend, from which point they are increased in thick¬ 
ness at the rate of 0.014 feet per foot to the base, or 1 in 72.5. 



Fig. 77. View of Big Bear Valley Dam with Old Dam Shown in Foreground 


The arc of the extrados of the arch ring is 140° 08' from the 
top to bottom the radius being maintained at 17 feet and the rise 
at 11.22 feet. The extrados is, therefore, a cylindrical surface 
uniform throughout, all changes in dimensions being made on the 
intrados of the arch. Thus at the top, the radius of the intrados 
is 16 feet, the arc 145° 08', and the rise 11.74 feet. At 80 feet from 
the top, Fig. 79, the thickness of the arch ring will be 2.15 feet, 
the radius of the intrados 14.85 feet (the radius of extrados less 
the thickness of the wall), the arc 140° 48' and the rise 10.59 





DAMS AND WEIRS 


133 

feet. In all cases of arch-dam design the clear span, radius, and 
rise of the intrados decrease from the top downward. 


m 

V 

<X> 

Sh 

« 

T5 

d 

d 

m 

<D 

-d 

Q 

"a 


a 

o z 
o $ 

I * 

m 


a 

o 

O 




V. 
<u 

be s 
a -2 

'% a 

O 6, 

C3 co 

Q £ 

3? § 

=3 O 
d 


c3 

<D 

ffl 

bO 

• H 

CQ 

o 

& 

a> 


M 

% 

« 


CO 

bb 

• rH 


Strut-tie members are provided between the buttresses to 
stiffen and take up any lateral thrusts that might be set up b\ 
seismic disturbances or vibrations, these consisting of T-beams 




I , fimmmm 1 


. 


i: - 

s 

f 1MBB ' 












134 


DAMS AND WEIRS 


and supporting arches all tied together by heavy steel reinforcement. 
The T-beams are 12 inches thick and 2.5 feet wide, with a 12- 
inch stem, set on an arch 12 inches square at the crown and thick¬ 
ening to 15 inches toward the springing lines, with two spandrel 
posts on each side connecting the beam and arch, all united into 
one piece. There are provided copings for the arches and the tops 
of the buttresses with 9-inch projections, making the arch cope 
2.5 feet wide and that on top of the buttresses 3 feet wide. The 
beam slab of the top strut members is built 4 feet wide to serve 
as an extra stiffener, as well as a comfortable footwalk across the 
dam. This footwalk is provided with a cable railing on both sides 



to make it a safe place upon which to walk. To add to the archi¬ 
tectural effect of the structure, the arches of the strut members 
terminate in imposts, built as part of the buttresses. The struts 
are reinforced with twisted steel rods, all being tied together and 
all being continuous through the buttress walls. The ends entering 
the buttresses are attached to other reinforcement passing cross¬ 
wise into the buttress walls, forming roots by which the stresses 
in the beams may be transmitted to and distributed in the buttress 
walls. The ends of the strut members are all tied onto the granite 
rock at both ends of the structure by hooking the reinforcement 
rods into drill holes in the rock. The buttresses are not reinforced, 
except to be tied to the arch rings and the strut members, their 



































DAMS AND WEIRS 


135 


shape and the loads they are to carry making reinforcement super¬ 
fluous. The arch ribs are reinforced with {-inch twisted rods hori¬ 
zontally disposed 2 inches from the inner surface and variably 
spaced. These rods were tied to the rods protruding from the 
buttresses. For reinforcing the extrados of the arch ring ribs of 
HXlJX^-inch angles were used, to which “ferro-inclave” sheets 
were clipped and used both as a concrete form for the outer face 
and a base for the plaster surface. 

104. Stress Analysis. On Fig. 79 a rough stress analysis 
is shown for 80 feet depth of water. As will be seen the resultant 
R cuts the base just short of the center point. The value of N 
is estimated at 4100 tons, the area of the base A = 110X4.2 = 460 

N 4100 

sq. feet whence —r = -—— =9 tons nearly, evenlv distributed (m being 

A 400 1 ° 


taken as unity). The stress on the arch, 80 feet deep, neglecting 
its weight is —-- - - = 20 tons, nearly. This shows the 

0 .jL /\ (jZj 


1 3 

necessity for the reinforcement provided to take — or of this stress. 

2 8 


P 3200 

The tangent of 0 = — = —— = .78. .’. 0 = 39°. 

N 4100 

This is a large value, 35 degrees being the usual limit, 33 degrees 
better. If the arch thickness were doubled, reinforcement would not 
be necessary except near the crest and the additional load of about 
320 tons would bring 6 down to 35 degrees. If not, a greater inclination 
given to the arch would increase the load of water on the extrados. 
It is quite possible that a thicker arch without reinforcement would 
be actually cheaper. The downward thrust acting on the arch 
due to its own weight is on a different plane from the arch thrust. 
Its effect is to increase the unit stress to a certain extent, as is also 
the case with the combination of shearing and compressive stresses 
in the interior of a dam as explained in Part I. This increase can, 
however, be neglected. A considerable but undefined proportion 
of the water pressure near the base is conveyed to it and not to 
the buttresses; this will more than compensate for any increase 
due to vertical compression and consequently it can be ignored. 
The ribs connecting the buttresses form an excellent provision for 
stiffening them against buckling and vibration and are universally 






13C 


DAMS AND WEIRS 


employed in hollow concrete dams. The buttresses in this instance 
are not reinforced. 

HOLLOW SLAB BUTTRESS DAMS 

105. Description of Type. There is a class of dam and weir 
similar in its main principles to the arch buttress type which is 
believed to have been first introduced by the Ambursen Hydraulic 
Construction Company of Boston. In place of the arch an inclined 
flat deck is substituted, which has necessarily to be made of rein¬ 
forced concrete. For this reason, the deck slabs cannot exceed a 
moderate width, so numerous narrow piers take the place of the 
thick buttresses in the former type. A further development is a 
thin deck which covers the downstream ends of the buttresses or 
piers, forming a rollway. The enclosed box thus formed is occa¬ 
sionally utilized as a power house for the installation of turbines, 
for which purpose it is well suited. 

The inclination given to the flat deck is such that the incidence 
of the resultant (R.F.) will fall as near the center of the base as pos¬ 
sible and at the same time regulate the inclination of the resultant 
to an angle not greater than that of the angle of friction of the 
material, i.e., 30 degrees with the vertical. By this means any 
tendency to slide on the foundation is obviated. 

Ellsworth Dam an Example. A good example of this style of 
construction is given in Fig. 80 of the Ellsworth dam in Maine. In 
this design the inclination of the deck is 45° or very nearly so; the 
piers are 15 feet centers with widened ends, so that the clear span 
of the concrete slabs is 9' 1" at the bottom. 

The calculations necessary to analyze the thickness of the 
slabs and the steel reinforcement at one point, viz, at El. 2.5, will 
now be given. In this case the pressure of water on a strip of the 
slab, one foot wide, the unsupported span of which is 9' 1", is IIlw. 
Here 77 = 67 feet and tv is ^ ton per cubic foot; therefore, W = 
67X9.1X^=19 tons. To this must be added the weight of the 
slab. As this latter lies at an angle with the horizontal its weight 
is partly carried by the base and is not entirely supported by the 
piers. The diagram in Fig. 80c is the triangle of forces. The weight 
of slab w is resolved in two directions, a and h, respectively, parallel 


DAMS AND WEIRS 


137 


and normal to face of slab. The angle being 45 degrees, a = 6=-^=* 

V 2 

Consequently the thickness, 37 inches, can be considered as reduced 


_ StJO-L _ 

oSot QOS' 0 



to 


3.1 

1.4 


= 2.2 feet. The 


portion of the weight of 
the slab carried to the 
piers will, therefore, be 

9.1 X2.2X-^r = 1.5 tons, 
40 

the weight of the con¬ 
crete being assumed at 
the usual value of 150 
pounds per cubic foot. 
The total distributed 
load in the strip will then 
be 19 + 1.5 = 20.5 tons. 

Now the moment of 
stress on a uniformly 
loaded beam with free 
Wl 


ends 


is 


8 


or 


M = 


20.5X109 

8 


= 279 inch- 


tons. 

This moment must 
be equaled by that of 
the resistance of the 
concrete slab. 

106. Formulas for 
Reinforced Concrete. 
For the purpose of 
sho-wing the calcula¬ 
tions in detail, some 
leading formulas con¬ 
nected with reinforced 
beams and slabs will 
now be exhibited: 











































































138 


DAMS AND WEIRS 



bd*- M ° 

fsVJ 

(26) 

or, approximately, 

IfsP 

(26a) 


u UM 

(27) 

or, approximately, 

bd * = f{ c 

6 Jc 

(27a) 


From these are found d, the required thickness of a slab up to 
centroid of steel, or M s -\-M c the bending moments, in which b is 
width of beam in inches; d depth of centroid of steel below top 
of beam; M c and M s symbolize the moments of resistance of the 
concrete and steel, respectively; f s safe unit fiber stress in steel, 
12,000 to 16,000 lb., or 6 to 8 tons per square inch;/ c safe extreme 
stress in concrete 500, 600, or 650 lb., or .25, .3, or .325 tofi per square 

inch; p steel ratio, or — 

bd 


n 


Ideal steel ratio p = - —- (28) 

^ 2(r 2 +rn) v y 

A area of cross-section of steel; k ratio of depth of neutral axis below 
top to depth of beam 

k = V 2 pn + (pn) 2 — pn (29) 


j ratio of arm of resisting couple to d 

j=( i-m 

E, 


(30) 


n ratio E s and E c being the moduli of elasticity, ordinary values 
E c 

fs A 

12 to 15; r ratio . As p = r~r, when reinforced slabs are analyzed, 

Jc bd 

formulas (26) and (27) can be transposed as below. 


From (26) M s =f s Ajd 

(31) 

From (26a) M s = \f s Ad Approximate 

(31a) 

From (27) M c = \f c kjbd 2 

(32) 

From (27a) M c Approximate 

(32a) 


In the case under review the reinforcement consists of three 
one inch square steel rods in each foot width of the slab. Using the 









DAMS AND WEIRS 


139 


approximate formulas (31a) and (32a), f s = 8 tons, / c = .3 ton, 
d = 35 inches and 6 = 12 inches; then 


M s = 8X3 X-^-X35 = 735 inch-tons 
8 

M c = ~X^rX 420X35 = 735 inch-tons 
1U o 


the results being identical. As already noted the moment of stress 
is but 279 inch-tons. The end shear may have governed the thick¬ 
ness. Testing for shear the load on a 12-inch strip of slab is 20.5 
tons of which one-half is supported at each end. Allowing 50 lb., or 
.025 ton, as a safe stress, the area of concrete required is 10.25 -f- .025 
= 410 square inches the actual area being 37 X 12 = 444 square inches. 

107. Steel in Fore Slope. The reinforcement of the fore slope 
is more a matter of judgment than of calculation, this deck having 
hardly any weight to support, as the falling water will shoot clear 
of it. The piers are not reinforced at all, nor is it necessary, as the 
stresses are all compressive and the inclination of the upstream deck 
is such that the resultant pressure makes an angle with the vertical 
not greater than that of friction, i.e., 30 degrees. , Fig. 80a is a force 
diagram of the resultant forces acting on the base at El. 0.00. The 
total weight of a 15-foot bay is estimated at 783 tons while that of P, 
the trapezoid of water pressure, is 1700 tons. The force line P in Fig. 
80 drawn through the c.g. of the water pressure area intersects the 
vertical force W below the base line. From this intersection R is 
drawn upward parallel to its reciprocal in the force polygon, cutting 
the base at a point some 9 feet distant from the center point. 

The maximum stress will occur at the heel of the base. A = 

0 01 . , N 2000 no . + i 107+54 

107X2 = 214 sq. ft.; -—= —— =9.34 tons; q being 9 ft., ra = ———— 

A 214 107 

= 1.5 and 5 = 9.34X1.5 = 14 tons. Formula (9), Part I. The hori¬ 
zontal component of P = 1200 tons. The base being 2 ft. wide, 


1200 

2X107 


= 5.6 


tons; 


therefore by formula (10), Part I, c = 7 + 


V49 + 31.4 = 16.5 tons, a decidedly high value. The usual limit to 
shearing stress is 100 lb. per sq. inch, equivalent to 7.2 tons per 
sq. ft., reinforcement is therefore not necessary and is not provided. 

There appears to be no reason why a steeper slope should not 
have been given to the deck so as to bring the center of pressure up 






140 


DAMS AND WEIRS 


to the center of the base and thus reduce the unit stress. Possibly a 
higher river stage has been allowed for. The position of 11 as well 
as the weight of the structure were obtained from the section given 
in Schuyler’s Reservoirs. Fig. 80 is of the so-termed “Curtain” 
type of dam. The “Half Apron” type, Fig. 82c, is sometimes used for 
overfalls, the main section of Fig. 82 illustrating the “Bulkhead ’ type. 

108. Slab Deck Compared with Arch Deck Dam. The Ambur- 
sen dam, wherever the interior space is not required for installa¬ 
tion of turbines, is undoubtedly a more expensive construction than 
the multiple arch type. This fact has at last been recognized and in 
one of the latest dams erected, scallop arches were substituted for 
the flat deck, thus obviating the expense of reinforcement. By 



increasing the width of the spans, the piers, being thicker in like 
proportion, will be in much better position for resisting compressive 
stress, as a thick column can stand a greater unit stress than a thin 
one. Another point in favor of the arch is that the effective length 
of the base of the piers extends practically to the crown of the arch. 
The arch itself need not be as thick as the slab. Owing to the liquid 
radial pressure to which it is subjected it is in a permanent state of 
compression and does not require any reinforcement except possibly 
at the top of the dam. Here the arch is generally widened, as in 
the case of the Ogden dam, Fig. 73, and thus greatly stiffened at the 
point where temperature variations might develop unforeseen stresses. 

Fig. 81 is a sketch illustrative of the saving in material afforded 
by doubling the spans from 15 to 30 feet and conversion to multiple 
arch type. The radius of the extrados of the arches is 18.5 ft. H 
is 67 at elevation 2.50 and w = ton; hence the thickness of the 
arch by formula ( 21 ) ($i being taken as 15 tons), will be 


b = 


RHw 18.5X67X1 


= 2.6 feet 


s 


32X15 










DAMS AND WEIRS 


141 


It is thus actually thinner than the reinforced slab of one-half 
the span, or 15 feet. Ihe greater length of the arch ring over that 
of the straight slab is thus more than compensated. The area of 
the arch, counting from the center of the pier, is 35X2.6 = 91 square, 
feet, that of the slab is 30X3.1 =93, that of the bracketing at junc¬ 
tion with the piers, 13, giving a total of 106 square feet. The saving 



due to decreased length of the piers is 25 square feet. Thus in the 
lower part of the dam over 40 cubic feet per 30' bay per foot in height 
of concrete is saved, also all the steel reinforcement. If a rollway 
is considered necessary in the weir, the deck could be formed by a 
thin reinforced concrete screen supported on I-beams stretching 
across between the piers. 

109. Guayabal Dam. Fig. 82 is a section of the Guayabal 
dam recently constructed in Porto Rico, its height is 127 feet and 
it is on a rock foundation. The following are the conditions govern- 



































































142 


DAMS AND WEIRS 


ing the design; maximum pressure on foundation 10 tons per square 
foot; compression in buttresses 300 pounds per square inch or 21.6 
tons per square foot; shear in buttresses 100 pounds per square 
inch, or 7.2 tons per square foot; shear in deck slabs 60 pounds, or 
.03 ton per square inch; f c for deck slabs 600 pounds or .3 ton per 
square inch; f s for deck slabs 14,000 pounds, or 7 tons per square 
inch. 

The concrete in the slabs is in the proportion of 1:2:4, in the 

E s . f s 

buttresses 1:3:6; n = —' is taken as 15 and r=+ = 23.3. The deck 

•Ac Jc 

slab is 55 inches thick at El. 224, d is taken as 53, allowing 2 inches 
for covering the steel, bd or the area of the section one foot wide = 
53 X12 = 636 square inches. Now A the area of the steel = pbd. By 

n 15 


formula (28), p = - 


= .01044, hence the 


2 (r 2 +m) 2(23.3) 2 +23.3 X15 

required area of steel will be 636X.01044 = 6.64 square inches, 

provided d is of the correct value. The calculation will now be 

made [for the thickness of the slab which is actually 55 inches. 

The load on a strip 12 inches wide is 

w , 109X13 .. 04 

Water pressure ——— = 44.3 tons 

To this must be added a portion of the weight of the slab 
which latter amounts to X = 4.5 tons. Of this 


45 

V2 


= 3.2 tons must be added to the 44.3 tons above, 44.3+3.2 = 47.5 


, rp, , 1 . , . WL 47.5X13X12 n<vy . . 

tons, the bending moment M is-=---= 927 mch- 

8 8 

tons. The depth of the slab can be estimated by using formulas 
(26) or (27) or the approximate ones (26a) and (27a). For the 
purpose of illustration, all four will be worked out. First the values 
of k and j will be found by formulas (29) and (30). 

= V.313 + .0245 -. 156 = .582 - .156 = 0.426 
j = (l=-D = l-.142 = .858 


By formula (26), d 2 = 




927 


12 f s pj 12X7X.0104X.858 


-= 1234 












DAMS AND WEIRS 


143 


d = V1234 = 35.07 inches 
By formula (27), d 2 = 


2 M r 927X2 


= 1406 


12 fjej 12 X.3 X.426 X.858 
d = 1406 = 37.5 inches 

Now the approximate formulas will be used. By (26a) 

8X927 1854 


d 2 = 


7X12X7X.0104 1.53 


= 1210 


V ; :T 


d = Vl210 = 34.8 inches 


by (27a) 


d 2 = 6X92 7 = 5562 = 

12 X.3 3.6 

d = Vl542 = 39.3 inches 


The approximate formulas (26a) and (27a) give higher results than 
(26) and (27) . The result to select is 37.5 inches, formula (27) , which is 
higher than by (26) . The depth of beam would then be 40 or 41 inches. 
It is actually 55. This discrepancy may be due to the water having 
been given a s.g. in excess of unity, owing to the presence of mud 
in suspension, say of 1.3 or 1.5, or shear is the criterion. 

The corresponding steel area will be A =pbd = .0104x12x37.5 
= 4.7 square inches. 1 te " round rods spaced 3 inches would answer. 
With regard to direct shear on the slab, W as before =47.5 tons of 
which half acts at each pier, viz, 23.7 tons. The safe resistance is 

23 7 

6dX$ s = 12X55X.03 = 20 tons, nearly. The shear = ’ ‘ — = .036 

12X55 

ton = 72 pounds per square inch. This figure exceeds the limit of 
60 pounds. The deficiency is made up by adding the shear of the 
steel rods. The sectional area of this reinforcement is 4.7 square 
inches the safe shearing of which is over 20 tons. These rods are 
usually turned up at their ends in order to care for the shear. 

Shear in Buttresses. With regard to shear in the buttresses, the 
horizontal component of the water pressure as marked on the force 
diagram is 3400 tons. The area of the base of the buttress at 

Q40f) 

El. 224 is 138X3.2 = 441.6, the shearing stress or s s then =77 tt; = 8 

441.6 

tons per square foot, nearly. The allowable stress being only 7.2 
tons the difference will have to be made good by reinforcing rods 
of which two of f-inch diameter would suffice. 













144 


DAMS AND WEIRS 



Now with regard to compressive stresses on the buttresses the 
graphical working shows that the resultant R strikes the base at El. 

224 almost exactly at the center, the angle 
6 also is 30 degrees. The value of N is 
5650 tons; si the mean and s the maximum 

N 

stress will both equal ; and A, the area 

of the base, equals 138X3.2 = 442 sq. ft.; 

therefore, s = —— = 12.78 tons. The corn- 
442 

pression on the foundation itself, which 
is 4 feet lower will not be any less for, 
although the base width is greater, N as 
well as P are also increased. Thus the 
pressure on the foundation is in excess 
of the limit and widening to a further 
extent is required. 

The maximum internal stress c, in 
the buttress at El. 224, will be by formula 


+3 


00 

tH 

o 

U-t 

O 

A 

bO 

a 

a -2 

ej — 

fi 5 

*3 -2 

ri c3 

.2 M 
•u (3 

o .a 
m a 

a 

s 

2 ® 
X i- 

0 ) 

" -t-i 

3 a 

M g 
° o 

fc -2 

•- ob 
> -< 

£ 5 

<U -r* 
& * 
m 

. 

CO o3 

oo a 
. ® 

- i-t r** 

fe “5 


^^-+5 S 2 . Here s = 12.8 


( 10 ), Part I, %s+- 
and s s as we have seen is 8 tons, therefore, 


164 


c = 6.4+^ —^-+64 = 16.6 tons. 


The limit 


in 

e 3 

X 

a 

o 

• 4-3 

o 

a> 

CO 

CO 

X) 

H 


compression in the buttress is 300 pounds 
per square inch, or 21.6 tons per square 
foot. 

In the bulkhead portion of the dam, 
shown in Fig. 82b, every pier is run up 
14 inches thick through the deck to form 
a support for a highway bridge, the spans 
of which are therefore 16 feet 10 inches 
in the clear; the roadway is carried on 
slabs which are supported by arches of 
reinforced concrete. The buttresses are 
laterally supported by several double rein¬ 
forced sway beams, 16"X14", and below the crest a through road¬ 
way is provided. The spillway section is shown on Fig. 82c. The 













DAMS AND WEIRS 


145 



Fig. 84. View of Bassano Dam over the Bow River Taken Just after Water Was Turned into Can 











































































146 


DAMS AND WEIRS 


ground level is here on a high bench at EL 295. The crest being 
EL 325, the fall is 30 feet. The spillway is of the “half apron type”. 
The roadway here is carried on four reinforced concrete girders, a 
very neat construction; the piers are run up every alternate span 
and are therefore at 36-foot centers; they are beveled on both faces 
to reduce end contraction. The spillway will pass 70,000 second- 
feet; its length is 775 feet. 

The bulkhead section of the dam (see also Fig. 83) has 51 spans 
of 18-foot centers, total length 918 feet; that of the spillway consists 
of 21 spans of 36-foot centers. The whole length is 1674 feet. The 
depth of the tail water is not known, it would probably be about 
20 feet and its effect would be but trifling. This is one of the largest 
hollow dams ever constructed. The arrangement of the haunches 
or corbels of the buttresses is a better one than that in the older 
work of Fig. 80. 

110. Bassano Dam. Another important work is the Bassano 
dam illustrated in Figs. 84 and 85. This is an overfall dam built over 
the Bow River at the head of the eastern section of the Canadian 
Pacific Railway Company’s irrigation canal and is estimated to pass 
100,000 second-feet of water at a depth of 14 feet. Though not so 
high nor so long as the Guayabal dam it presents several features of 
interest. First its foundations are on a thick blanket of clay some 
twelve feet deep which overlies boulders and gravel. This material 
is very hard blue clay of excellent quality. The great advantage of 
this formation, which extends over 1000 feet upstream from the work, 
is that it precludes all uplift, or very nearly so, consequently no 
special precautions have to be adopted, such as a long apron to 
ensure length of percolation, as would be necessary in case of a 
foundation composed of porous and loose materials. It has also 
disadvantages. The allowable pressure on the clay is limited to 
2\ tons per square foot. This influences the design necessitating 
a wide spread to the buttresses, laterally as well as longitudinally. 
The whole of the dam is an overfall and the general arrangements 
are very similar to those prevailing at Guayabal. The hearth or 
horizontal fore apron, a provision not necessary in the last example, 
is at EL 2512. The crest is at 2549.6 a height of 37.6 feet above 
the apron and corresponds with the level of the canal intake floor. 
Water is held up to eleven feet above crest level by draw gates 


DAMS AND WEIRS 


147 



























































































































































































































































































148 


DAMS AND WEIRS 


eleven feet high, and this full supply level is three feet below that of 
the estimated afflux, which is fourteen feet above the crest. 

For overturning moment the water-pressure area will be a 
truncated triangle with its apex at afflux level plus the height h or 

1.5 — to allow for velocity of approach, as explained in section 
2g 

57, Part I. This, in the Bow River with a steep boulder bed will be 

144 

about 12 feet per second; h therefore will equal 1.5 X-tt =3.4 feet and 

64 

the apex of the truncated triangle will be at a point 14+3.4 = 17.4 
feet above the crest level. The depth of the tail water at full flood 

is not known, the ratio ~ with a steep bed slope will not be under .5, 

consequently with d= 14, D will have a value of about 25 to 28 feet, 
d being depth over crest and D that of tail water. The overturning 
moments direct and reverse can be represented by the cubes of the 
depths up- and downstream and the unbalanced moment by their 
difference. The upstream head is 37.0+14+3.4 = 55 feet and the 
downstream head say 25 feet. Their cubes are 166,375 and 15,625 
the difference being 150,750, thus the reverse pressure will not have 
much effect in assisting the stability of the structure. The cor¬ 
responding representative moment when water is held up to 11 ft. 
above crest will be 49 3 = 117,649, supposing the tail channel empty. 
This quantity is less than the 150,750 previously stated, consequently 
the afflux level is that which has to be considered when estimating 
the overturning moment. In the case of direct water pressure on 
the deck slabs, the acting head at full flood will be the difference of 
the flood level up- and downstream, which is 30 feet, as the tail 
water is allowed access to the rear of the deck slabs. This is less 
than the head, 1 ’ 49 feet, which exists when the gates are closed and 
water is held up to canal full supply, i.e., to EL 2560.6, consequently 
the head that has to be considered is that at this latter stage. 

Analysis of Pressures on Bassano Dam. With this data the 
design can be analyzed, the procedure being identical with that 
explained in the last example, excepting that the reverse pressure 
might be taken into account as it will modify the direction and 
incidence of R in a favorable sense though not to any great extent. 
The limit stresses are those given in the last example with the fol- 



DAMS AND WEIRS 


149 


lowing additions: Footings, compression in bending, 600 pounds 
per square inch, shear, 75 pounds per square inch. 

Some explanation will now be given of the method of calcu¬ 
lation of the footing to the buttresses and the Guayabal dam will 
be referred to, as the pressures on the base of the buttresses are 
known quantities. In section 109 the value of N is 5650 tons and 


N 5650 


= 41 tons, nearly. This is the unit pressure per foot run 


b 138 

on the base of the buttress. Supposing the limit pressure on the 
foundation was fixed at 3 tons per square foot, then the requisite 

41 

base width of the footing would be — = 13.7 feet. The footing con- 

O 


sists of two cantilevers attached to the stem of the buttress. The 
bending moment M at the junction with the buttress of a strip 1 


foot wide will be 


Wl 

2 * 


The buttress being 3.2 feet wide the pro¬ 


jecting length of footing on each side will be 


13.7-3.2 

2 


= 5.25 feet. 


The reaction on a strip one foot wide will be 5.25X3X1 = 

15.75 tons. The moment in inch-tons about the edge of the section 

, +1 , ., | 12 Wl 15.75X5.25X12 , no . . 

or the buttress will be —— =---=498 mch-tons. 

— 1 Urn 

According to formula (27a), bd 2 = ^~. .'. d= = V830 = 

jc * 12 X .5 


28.8 inches. Then bd = 28.8X12 = 346 and A the area of the steel 
at the base will be pbd = .0104X346 = 3.61 inches, this in a 12-inch 
wide strip will take lj-inch bars 4 inches apart. When the weight 
on the buttress is considerable the depth of footing slab thus esti¬ 
mated becomes too great for convenience. In such cases, as in Fig. 
85, the beam will require reinforcement in compression at the top. 
This complicates the calculation and cases of double reinforcement 
are best worked out by means of tables prepared for the purpose. 
The footings shown in Fig. 85, were thus double reinforced, in fact 
through bars were inserted at each step, the lower being in tension 
the upper ones in compression. The lower bars were continuous 
right through the base of the dam. This reinforcement of the 
footing is not shown on the blue print from which Fig. 85 is derived. 

111. Pressure on Foundation Foredeck. A great many 











150 


DAMS AND WEIRS 


Ambursen dams have been constructed on river beds composed 
of boulders and gravel, which require a pressure limit of about 
4 tons per square foot. This can always be arranged for by widen¬ 
ing the footing of the pier buttresses, the same can of course be 
done with arch buttressed dams. The base of the arch itself can 
be stepped out in a similar manner. In the Bassano dam the 
sloping fore deck is unusually thick and is heavily reinforced in 
addition; this is done with the idea of strengthening the structure 
against shock from ice, as well as from the falling water, and with 
the further idea of assisting the buttresses in carrying the heavy 
load of the piers and superstructure. It is doubtful if any calcu¬ 
lations can well be made for this; it is a matter more of judgment 
than of estimation. 

Buttresses, As with the Guayabal spillway, every alternate 
buttress is run up to form the piers of the superstructure, which latter 
consists of a through bridge which carries the lift gear for manipu¬ 
lating the draw gates. The so-termed blind buttresses—that is, 
those that do not carry a pier—are of thinner section and are appar¬ 
ently not reinforced. Both kinds of buttresses have cross-bracing 
as shown on the profile. In hollow dams the location of the center 
of pressure moves with the rise of water from the heel toward the 
center within the upstream half of the middle third. In solid dams, 
on the other hand, the movement is along the whole of the middle 
third division, consequently in hollow dams there is no tendency 
to turn about the toe as with solid dams, rather the reverse, namely, 
to upset backward. This latter tendency must cause tension in 
the buttresses which the cross-bracing is intended to care for. 

Baffles. As noted already in section 66, baffles have been built 
on the curved bucket with the object of neutralizing this mischiev¬ 
ous arrangement which it is hoped will soon [become as obsolete in 
western practice as has long been the case in the East. 

Hearth and Anchored Apron. The dam is provided with a solid 
horizontal fore apron or hearth 76 feet long and beyond this the 
device of an anchored floating apron of timber 30 feet in length has 
been added. The apron is undoubtedly too short and should have 
been made 100 feet or 2 (77+d) in length, with cribbed riprap 
below it. The wooden sheet piling in the rear of the work is con¬ 
sidered to be worse than useless; it merely breaks up the good clay 


DAMS AND WEIRS 


• 151 


blanket by cutting it in two. A wide solid curtain of concrete, not 
so deep as to penetrate the clay blanket, would have been a superior 
arrangement. The inclined piling below the bucket is provided to 
guard against sliding. This dam is provided with a number of sluice 
openings. Their capacity is such that one half will pass ordinary 
floods, allowing the other half of the dam to be cut off from the river 
by sheet piling during construction. On completion of the work 
these sluices were all closed from the inside by slabs of concrete 
deposited in position. 

SUBMERGED WEIRS FOUNDED ON SAND 

112. Description of Type. There is a certain type of drowned 
or submerged diversion weir which is built across wide rivers or 
streams whose beds are composed of sand of such depth that a solid 
foundation on clay is an impossibility. Consequently, the weir has 
to be founded on nothing better than the surface of the river bed, 
with perhaps a few lines of hollow curtain walls as an adjunct. Of 
this class of weir but one is believed to have been constructed in 
the United States, viz, the Laguna weir over the Colorado River 
at the head of the Yuma irrigation canals. 

This type originated in India and in that country are found 
numerous examples of weirs successfully constructed across very 
large rivers of immense flood discharge. For instance, the Goda- 
veri River in Southern India has a flood discharge of 1,200,000 
second-feet and the weirs across it are nearly 2\ miles in length. 
Not only is the length great, but as will be seen, the width has to be 
very considerable. The Okhla weir, Figs. 101 and 102, situated on 
the Jumna below the historic city of Delhi is 250 feet wide and 
f mile long. The height of these submerged weirs is seldom over 
12 feet, their role being purely diversion, not storage. No doubt 
more of this type of low diversion weirs will in the future have to be 
constructed in the United States or in Mexico, so that a knowledge 
of the subject is a necessity for the irrigation engineer. 

Principles of Design. The principles underlying the successful 
design of these works are a comparatively recent discovery. Designs 
were formerly made on no fixed principles, being but more or less 
modified copies of older works. Fortunately some of these works 
failed, and it is from the practical experience thus gained that a 


152 


DAMS AND WEIRS 


knowledge of the hydraulic principles involved has at last been 
acquired. 

A weir built on sand is exposed not only to the destructive 
influences of a large river in high flood which completely submerges 
it, but its foundation being sand, is liable to be undermined and 
worked out by the very small currents forced through the under¬ 
lying sand by the pressure of the water held up in its rear. In spite 
of these apparent difficulties, it is quite practicable to design a work 
of such outline as will successfully resist all these disintegrating 
influences, and remain as solid and permanent a structure as one 
founded on bed rock. 

113. Laws of Hydraulic Flow. The principle which underlies 
the action of water in a porous stratum of sand over which a heavy 
impervious weight is imposed is analogous to that which obtains in 



Fig. 86. Diagram Showing Action of Water Pipe Leading Out of Reservoir 


a pipe under pressure. Fig. 86 exemplifies the case with regard to 
a pipe line BC, leading out of a reservoir. The acting head (II) 
is the difference of levels between A i a point somewhat lower than 
A, the actual summit level and C the level of the tail water beyond 
the outlet of the pipe. The water having a free outlet at C , the 
line AiC is the hydraulic gradient or grade line. The hydrostatic 
pressure in the pipe at any point is measured by vertical ordinates 
drawn from the center of the pipe to the grade line AiC. The uni¬ 
form velocity of the water in the pipe is dependent directly on the 
head and inversely on the frictional resistance of the sides of the pipe, 
that is, on its length. This supposes the pipe to be straight, or 
nearly so. 

114. Percolation beneath Dam. We will now consider the 

case of an earthen embankment thrown across the sandy bed of a 































DAMS AND WEIRS 


153 


stream, Fig, 87. The pressure of the impounded water will natu¬ 
rally cause leakage beneath the impervious earthen base. With a 
low depth of water impounded it may well be understood that such 
leakage might be harmless; that is, the velocity of the percolating 
under current would be insufficient to wash out the particles of 
sand composing the foundation of the dam. When, however, the 
head is increased beyond a safe limit, the so-termed piping action 
will take place and continue until the dam is completely undermined. 

115. Governing Factor for Stability. The main governing 
factor in the stability of the sand foundation is evidently not the 
superimposed weight of the dam, as the sand is incompressible; 
although a load in excess of the hydraulic pressure must exercise 
a certain though possibly undefined salutary effect in delaying the 
disintegration of the substratum. However this may be, it is the 



Fig. 87. Diagram Showing Effect of Percolation under Earthen Embankment across Stream 


enforced length of percolation, or travel of the under current, that 
is now recognized to be the real determining influence. 

In the case of a pipe, the induced velocity is inversely propor¬ 
tional to the length. In the case under consideration, the hydraulic 
condition being practically identical with that in a pipe, it is the 
enforced percolation through the sand, and the resulting friction 
against its particles as the water forces its way through, that effects 
the reduction of the velocity of the undercurrent, and this frictional 
resistance is directly proportional to the length of passage. In the 
case of Fig. 87, the length of enforced percolation is clearly that of 
the impervious base of the earthen dam. The moment this obstruc¬ 
tion is passed the water is free to rise out of the sand and the hydro¬ 
static pressure ceases. 

116. Coefficient of Percolation. This length of enforced per¬ 
colation or travel, which will be symbolized by L, must be some 
































154 


DAMS AND WEIRS 


multiple of the head II, and if reliable safe values for this factor can 
be found, suitable to particular classes of sand, we shall be enabled 
to design any work on a sand foundation, with perfect confidence in 
its stability. If the percolation factor be symbolized by c, then L, or 
the length of enforced percolation, will equal ell, II being the head of 
water. The factor c will vary in value with the quality of the sand. 

Fig. 88 represents a case similar in every respect to the last 
except, instead of a dam of earth, the obstruction consists of a 
vertical wall termed the weir or drop wall, having a horizontal 
impervious floor attached thereto, which appendage is necessary to 
prevent erosion of the bed by the current of falling water. 

At the stage of maximum pressure the head water will be 
level with the crest, and the level of the tail water that of the floor; 
consequently the hydraulic gradient will be IIB, which is also the 
piezometric line and as in the previous case of the pipe line, Fig. 


H J 



86, the ordinates of the triangle IIAB will represent the upward 
hydrostatic pressure on the base of the weir wall and of the floor. 

117. Criterion for Safety of Structure. The safety of the 
structure is evidently dependent on the following points: 

First, the weir wall must be dimensioned to resist the overturn¬ 
ing moment of the horizontal water pressure. This has been dealt 
with in a previous section. Second, the thickness, i. e., the weight 
of the apron or floor must be such that it will be safe from being 
blown up or fractured by the hydrostatic pressure; third, the base 
length, or that of the enforced percolation L , must not be less than 
cH, the product of the factor c with the head IL Fourth, the 
length of the masonry apron and its continuation in riprap or con¬ 
crete book blocks must be sufficient to prevent erosion. 

It is evident that the value of this factor c , must vary with the 
nature of the sand substratum in accordance with its qualities of 















DAMS AND WEIRS 


155 


fineness or coarseness. Fine light sancl will be closer in texture, 
passing less water under a given head than a coarser variety, but at 
the same time will be disintegrated and washed out under less 
pressure. Reliable values for c, on which the design mainly depends, 
can only be obtained experimentally, not from artificial experiments, 
but by deduction from actual examples of weirs; among which the 
most valuable are the records of failures due to insufficiency in 
length of percolation. From these statistics a safe value of the 
relation of L to II, the factor c, which is also the sine of the 
hydraulic gradient, can be derived. 

118. Adopted Values of Percolation Coefficient. The follow¬ 
ing values of c have been adopted for the specified classes of sand. 

Class I: River beds of light silt and sand, of which 60 per cent 
passes a 100-mesh sieve, as those of the Nile or Mississippi; percola¬ 
tion factor c = 18. 

Class II: Fine micaceous sand of which 80 per cent of the 
grains pass a 75-mesh sieve, as in Himalayan rivers and in such as 
the Colorado; c — 15. 

Class III: Coarse-grained sands, as in Central and South 
India; c = 12. 

Class IV: Boulders or shingle and gravel and sand mixed; c 
varies from 9 to 5. 

In Fig. 88 if the sand extended only up to the level C, the length 
of percolation would be CD, the rise from D to B not being counted 
in. In that case the area of hydrostatic pressure acting beneath 
the floor would be the triangle IIAB. As, however, a layer of sand 
from A to C interposes, the length will be ACD, and outline HiB. 
The step H Hi occurring in the outline is due to the neutralization 
of head symbolized by h, effected in the depth AC. Supposing 
AC to be 6 feet and the percolation factor to be 12, then the step 
in the pressure area, equal to h, will be 6-h 12 = 6 inches. The 
resulting gradient IIiB will, however, be flatter than 1 in 12; conse¬ 
quently the termination of the apron can be shifted back to BiDi, 
IIiBi, being parallel to HB; in which case the area of hydrostatic 
pressure will be HiABi. The pressure at any point on the base is 
represented by the ordinates of the triangle or area of pressure. 
Thus the upward pressure at E, below the toe of the drop wall, 
where the horizontal apron commences, is represented by the line 


156 


DAMS AND WEIRS 


FG. Supposing the head IIA to be 10 feet, then the total required 
length of percolation will be cH= 12X10 = 120 feet. This is the 
length ACD\. The neutralization of head, h, effected by the enforced 
percolation between H and G is represented by GJ , and supposing 
the base width of the drop wall CE to be 9 feet, AC being 6 feet, 

= = 1J feet. The upward pressure FG is (H—h) = 10 —1J 

= 8f feet. 

The stepped upper line bounding the pressure area as has been 
noted in Part I, is termed the piezometric line, as it represents the 
level to which water would rise if pipes were inserted in the floor. 
It is evident from the above that when no vertical depressions 
occur in the line of travel that the piezometric line will coincide 
with the hydraulic gradient or virtual slope; when, however, vertical 
depressions exist, reciprocal steps occur in the piezometric line, 
which then falls below the hydraulic grade line. The piezometric 
line is naturally always parallel to the latter. The commencement 
of the floor at E is always a critical point in the design as the pres¬ 
sure is greatest here, diminishing to zero at the end. 

119. Simplifying the Computations. In the same way that 
the water pressure is represented by the head producing it, the 
common factor w, or the unit weight of water, may also be elimi¬ 
nated from the opposing weight of the floor. The weight of the 
masonry, therefore, is represented by its thickness in the same way 
as the pressure, and if t be the thickness of the floor, tp will represent 
its weight. Now the floor lies wholly below low water level. Con¬ 
sequently, in addition to the external hydrostatic pressure repre¬ 
sented by H , due to the head of water upheld, there is the buoyancy 
due to immersion. The actual pressure on the base CD i is really 
measured by HC , not IIA. Thus if a vertical pipe were inserted 
in the floor the w T ater would rise up to the piezometric line and be 
in depth the ordinate of the pressure area plus the thickness of the 
floor. But it is convenient to keep the hydrostatic external pres¬ 
sure distinct from the effect of immersion. This latter can be 
allowed for by reduction in the weight of these parts of the struc¬ 
ture that he below L. W. L. See sections 52 and 53, Part I. 

Effect of Immersion. When a body is immersed in a liquid it 
loses weight to the extent of the weight of the liquid displaced. 



DAMS AND WEIRS 


157 


Thus the unit weight of a solid is wp. When immersed, the unit 


weight will be w(p — 1 ). As 


w 



is a discarded factor, the unit 
weight being represented 
only by p, the specific 
gravity, the weight of 
the floor in question will 
be t(p — 1 ) if immersed. 
We have seen that the 
hydrostatic pressure act¬ 
ing at F is 8 | feet. To 
meet this the weight, or 
effective thickness, of 
the floor must be equal 
to 8 f feet of water +J 
for safety, or, in sym- 

11 / H ~ h v 4 

bols, t = -- X— 

p — 1 o 

Assuming a value for p 
of 2 , the thickness re¬ 
quired to counterbal¬ 
ance the hydrostatic 
pressure will be 

£ = 8 fX- 7 r = 11.6 feet 


a 

o 


o 

« 

h. 

4) 

P-l 


<D 

> 

c3 

H 

o 

43 

-t-s 

M 

(3 

0> 

t-i 


0 
cr 

<u 

P5 

M 

d 

• rH 

U 

'> 

O 

Ph 


"d 

o 

-4> 

4) 


M 

a 

I 

o 

43 

CO 

13 

S3 

u3 

CO 

S3 

O 

h 

• r 4 

£ 

h-l 

o 

s 

o 

Ih 

Ph 


The formula for 
thickness will then stand: 
4 /1I-K 
3 




(S) 


(33) 


05 

co 

bi 

s 


p-l 

Uplift on Fore 
Apron. It is evident 
that in Fig. 88 the long 
floor is subjected to a 
very considerable uplift 
measured by the area 
IIAB, the weight of the 
apron also is reduced in 
the ratio of p : (p-l) as it lies below L.W.L., consequently it will 
have to be made as already noted of a depth of 11.6 feet which is a 








































































































158 


DAMS AND WEIRS 


quite impossible figure. The remedy is either to make the floor porous 
in which case the hydraulic gradient will fall below 1 in 12, and failure 
will take place by piping, or else to reduce the effective head by the 
insertion of a rear apron or a vertical curtain wall as has been already 
mentioned in section 53, Part I. In these submerged weirs on 
large rivers and in fact in most overfall dams a solid fore apron is 
advisable. The length of this should however be limited to abso¬ 
lute requirements. This length of floor is a matter more of individual 
judgment or following successful precedent than one of precise 
estimation. 

The following empirical rule which takes into account the nature 
of the sand as well as the head of water is believed to be a good guide 
in determining the length of fore apron in a weir of this type, it is 

L=2,<7h (34) 


In the case of Fig. 89, the head is 10 feet and c is assumed at 12, conse¬ 
quently, L = 3Vl20 = 33 feet, say 36, or 3c. In Fig. 89 this length 
of floor has been inserted. Now a total length of percolation of 12 
times the head, or 120 feet = 10c is required by hypothesis, of this 
3c is used up by the floor leaving 7c to be provided by a rear apron 
and curtain. Supposing the curtain is made a depth equal to 1 \c, 
this will dispose of 3 out of the 7 (for reasons to be given later), 
leaving 4 to be provided for by the rear apron, the length of which, 
counting from the toe of the weir wall, is made 4c or 48 feet. The 
hydraulic gradient starts from the point IV which is vertically 
above that of ingress A. At the location of the vertical diaphragm 
of sheet piling, a step takes place owing to the sudden reduction of 
head of 3 feet, the obstruction being 3c in length counting both 
sides. From here on, the line is termed the piezometric line and the 
pressure area is the space enclosed between it and the floor. The 
actual pressure area would include the floor itself, but this has been 
already allowed for in reduction of weight, its s.g. being taken as 
unity instead of 2. 

The uplift on the weir wall is the area enclosed between its 
base and the piezometric line In calculating overturning moment, 
if this portion were considered as having lost weight by immersion 
it would not quite fully represent the loss of effective weight due to 
uplift, because above the floor level the profile of the weir wall is 











DAMS AND WEIRS 


159 


not rectangular, while that of the pressure area is more nearly so. 
The foundation could be treated this way, the superstructure above 
AF being given full s.g. and the uplift treated as a separate vertical 
force as was the case in Fig. 40, Part I. 

120. Vertical Obstruction to Percolation. Now with regard 
to the vertical obstruction, when water percolates under pressure 
beneath an impervious platform the particles are impelled upward 
by the hydrostatic pressure against the base of the dam and also 
there is a slow horizontal current downstream. The line of least 
resistance is along the surface of any solid in preference to a shorter 
course through the middle of the sand, consequently when a vertical 
obstruction as a curtain wall of masonry or a diaphragm of sheet 
piling is encountered the current of water is forced downward and 
the obstruction being passed it ascends the other side up to the base 
line which it again follows. The outer particles follow the lead 
of the inner as is shown by the arrows in Fig. 89. The value of a 
vertical obstruction is accordingly twice that of a similar horizontal 
length of base. Valuable 'corroboration of the reliability of the 
theory of percolation adopted, particularly with regard to reduction 
of head caused by vertical obstruction, has been received, while this 
article was on the press, from a paper in the proceeedings of the 
American Society of Civil Engineers entitled “The Action of Water 
under Dams” by J. B. T. Coleman, which appeared in August, 1915. 
The practical value of the experiments, however, is somewhat 
vitiated by the smallness of the scale of operations and the dispro¬ 
portion in the ratio II: L to actual conditions. The length of base 
of the dam experimented on should not be less than 50 feet with a 
head of 5 feet. 

121. Rear Apron. The extension of the floor rearward is 
termed the rear apron. Its statical condition is peculiar, not being 
subject to any upward hydrostatic pressure as is the case with the 
fore apron or floor. Inspection of the diagram, Fig. 89, will show 
that the water pressure acting below the floor is the trapezoid enclosed 
between the piezometric line and the floor level; whereas the down¬ 
ward pressure is represented by the rectangle II iA ill A, which is 
considerably larger. Theoretically no weight is required in the 
rear apron, the only proviso being that it must be impervious and 
have a water-tight connection with the weir wall, otherwise the 


160 


DAMS AND WEIRS 


incidence of H may fall between the rear apron and the rest of the 
work, rendering the former useless. Such a case has actually 
occurred. It is, however, considered that the rear apron must be 
of a definite weight, as otherwise the percolation of water under¬ 
neath it would partake of the nature of a surface flow, and so pre¬ 
vent any neutralization of head caused by friction in its passage 
through sand. Consequently, the effective thickness, or rather 
t (p — 1) should not be less than four feet. Its level need not be the 
same as that of the fore apron or floor. In fact, in some cases it 
has been constructed level or nearly so with the permanent crest 



Fig. 90. View of Grand Barrage over Nile River 


of the drop wall. But this disposition has the effect of reducing 
the coefficient of discharge over the weir and increasing the afflux 
or head water level, which is open to objection. The best position 
is undoubtedly level with the fore apron. 

Another point in favor of the rear apron is the fact that it is 
free from either hydrostatic pressure or the dynamic force of falling 
water, to which the fore apron is subject; it can, therefore, be con¬ 
structed of more inexpensive material. Clay consolidated when 
wet, i.e., puddle, is just as effective in this respect as the richest 
cement masonry or concrete, provided it is protected from scour 
where necessary by an overlay of paving or riprap, and has a reli- 







DAMS AND WEIRS 


161 


able connection with the drop wall and the rest of the work. In old 
works these properties of the rear apron were not understood, and 
the stanching of the loose stone rear apron commonly provided, was 
left to be effected by the natural deposit of silt. This deposit 
eventually does take place and is of the greatest value in increasing 
the statical stability of the weir, but the process takes time, and 
until complete, the work is liable to excess hydrostatic pressure and 
an insufficient length of enforced percolation, which would allow 
piping to take place and the foundation to be gradually undermined. 

122. First Demonstration of Rear Apron. The value of an 
impervious rear apron was first demonstrated in the repairs to the 
Grand Barrage over the Nile, Fig. 90, some time in the eighties. 
This old work was useless owing to the great leakage that took 
place whenever the gates were lowered and a head of water applied. 
In order to check this leakage, instead of driving sheet piling, which 



Fig. 91. Section Showing Repairs Made on the Grand Barrage 

it was feared would shake the foundations, an apron of cement 
masonry 240 feet wide and 3.28 feet thick, Fig. 91, was constructed 
over the old floor, extending upstream 82 feet beyond it. This 
proved completely successful. By means of pipes set in holes 
drilled in the piers, cement mortar was forced under pressure into 
all the interstices of the rubble foundations, filling up any hollows 
that existed, thus completely stanching the foundations. So effec¬ 
tually was the structure repaired that it was rendered capable of 
holding up about 13 feet of water; whereas, prior to reconstruction, 
it was unsafe with a head of a little over three feet. The total 
length of apron is 238 feet, of which 82 feet projects upstream 
beyond the original floor and 44 feet downstream, below the floor 
itself, the latter having a width of 112 feet. The head II being 

13 feet, and L being 238 feet, c, or the percolation factor is = 

18, which is the exact value assigned for Nile sand in Class I, 





































162 


DAMS AND WEIRS 


section 118. This value was not originally derived from the Grand 
Barrage, but from another work. 

The utility of this barrage has been further augmented by the 
construction of two subsidiary weirs below it, see Figs. 92 and 106, 
across the two branches of the Nile delta. These are ten feet high 
and enable an additional height of ten feet to be held up by 
the gates of the old barrage, the total height being now 22 \ feet. 
The increased rise in the tail water exactly compensates for the 
additional head on the work as regards hydrostatic pressure, but 
the moment of the water pressure on the base of the masonry piers 



Fig. 92. Plan of Grand Barrage over Nile River Showing Also Location 
of Damietta and Rosetta Weirs 


will be largely increased, viz, as from 13 3 to 22 3 —10 3 , or from 2197 
to 9648. 

The barrage, which is another word for “open dam” or bulk¬ 
head dam, is, however, of very solid and weighty construction, and 
after the complete renewal of all its weak points is now capable of 
safely enduring the increased stress put upon the superstructure. 
We have seen, in section 119, that the width of the impervious 

fore apron should be L = 3Vci/, formula (34). This width of the 
floor is affected by two considerations, first, the nature of the river 
bed, which can best be represented by its percolation factor c and 
second, by the height of the overfall including the crest shutters 
if any, which will be designated by II a to distinguish it from i/, 
which represents the difference between head and tail water and also 















DAMS AND WEIRS 


163 


TABLE I 


Showing Actual and Calculated Values of Li or Talus Width 

Formula (35), 


River 

Name of Work 

Type 

c 

H 


Length Lj 


b 

Q 

Calculated 

Actual 

Ganges 

Narora 

A 

15 

10 

75 

150 

140-170 

Coleroon 

Coleroon 

A 

12 

41 

100 

92 

72 

Vellar 

Pelandorai 

A 

9 

11 

100 

108 

101 

Tampra- 

parni 

Srivakantham 

A 

12 

6 

90 

102 

106 

Chenab 

Khanki 

B 

15 

7 

150 

182 

170 

Chenab 

Merala 

B 

15 

7 

150 

182 

203 

Jhelum 

Rasul 

B 

15 

6 

155 

160 

135 

Penner 

Adimapali 

B 

12 

81 

184 

172 

184 

Penner 

Nellore 

B 

12 

9 

300 

228 

232 

Penner 

Sangam 

B 

12 

10 

147 

168 

145 

Godaveri 

Dauleshwiram 

B 

12 

13 

100 

158 

217 

Jumna 

Okhla 

C r 

15 

10 

140 

210 

210 

Kistna 

Beswada 

c 

12 

13 

223 

236 

220 

Son 

Dehri 

c 

12 

8 

66 

100 

96 

Mahanadi 

Jobra 

c 

12 

100 

140 

163 

143 

Madaya 

Madaya 

c 

12 

8 

280 

207 

235 

Colorado 

Laguna 

c 

15 

10 

1 

below 

minimum 

140 

200 


Type A has a direct overfall, with horizontal floor at L. W. L., as in 
Figs. 91, 93, and 95. 

Type B has breast wall followed by a sloping impervious apron, Figs. 
96 and 97. 

Type C has breast wall followed by pervious rock fill, with sloping surface 
and vertical body walls, Figs. 101 to 106. 

from II b the height of the permanent crest above L. W. L. Tak¬ 
ing the Narora weir as standard, a length of floor equal to 3 Vci/ 
= 3Vl5Xl3 = 42 feet, is deemed to be the correct safe width for 

























164 


DAMS AND WEIRS 


a weir 13 feet in height; where the height is more or less, the 
width should be increased or reduced in proportion to the square 
root of the height and that of the factor c. 

123. Riprap to Protect Apron. Beyond the impervious floor 
a long continuation of riprap or packed stone pitching is required. 
The width of this material is clearly independent of that appro¬ 
priate to the floor, and consequently will be measured from the 
same starting point as the floor, viz, from the toe of the drop wall. 
The formula for overfall weirs is 

Ll ~" k 'Jw x 'ln < 35 > 

For sloping aprons, type B, the coefficient of c will be 11 
Then 

cm.) 

This formula is founded on the theory that the distance of the 
toe of the talus from the overfall will vary with the square root of 
the height of the obstruction above low water, designated by II b, 
with the square root of the unit flood discharge over the weir crest 
q, and directly with c, the percolation factor of the river sand. The 
standard being these values; viz, 10, 75, and 10, respectively, in 
Narora weir. This height, II b is equal to H when there are no 
crest shutters, and is always the depth of L. W. L. below the per¬ 
manent masonry crest of the weir. This formula, though more or 
less empirical, gives results remarkably in consonance with actual 
value, and will, it is believed, form a valuable guide to design. 
Table I will conclusively prove this. As nearly all the weirs of 
this class have been constructed in India, works in that country 
are quoted as examples. 

124. Example of Design Type A. Another example of design 
in type A will now be given, Fig. 93, the dimensions being those of 
an actual work, viz, the Narora weir over the Ganges River, the 
design being thus an alternative for that work, the existing section 
of which is shown in Fig. 95 and discussed in section 125. The 
data on which the design is based, is as follows: sand, class 2; per¬ 
colation factor c = 15; II or difference between head and low water, 
the latter being always symbolized by L. W. L., 13 feet, unit dis- 





DAMS AND WEIRS 


1G5 


§ 



c3 

Ph 

<V 

> 

o 

+3 

o 

o 

5 



o 


'm 

o 

p 


CO 

Gb 

bi 

S 


charge over weir q = 75 second-feet, the 
total length of the impervious apron and 
vertical obstructions will, therefore, have 
to be L = c77= 15X13 = 195 feet. 

The first point to be determined is 
the length of the floor or fore apron. 
Having fixed this length, the balance of 
L will have to be divided among the 
rear apron and the vertical sheet piling. 
It is essential that this minimum length 
be not exceeded, as it is clearly of advan¬ 
tage to put as much of the length into 
the rear apron as possible, owing to the 
inexpensive nature of the material of 
which it can be constructed. According 
to formula (2), L = 42 feet, which is nearly 
equivalent to 3c, or 45 feet, there thus 
remains 10c to be proportioned between 
the rear apron and the vertical curtain. 
If the latter be given a depth of 2c, or 30 
feet the length of travel down and up will 
absorb 4c, leaving 6c, or 90 feet for the 
rear apron. The measurement is taken 
from the toe of the drop wall. The 
neutralization of the whole head of 13 
feet is thus accomplished. A second cur¬ 
tain will generally be desirable at the 
extremity of the fore apron as a pre¬ 
cautionary measure to form a protec- 
tion in case the loose riprap downstream 
from the apron is washed out or sinks. 
This curtain must have open joints to 
offer as little obstruction to percolation 
as possible. The outline of the pressure 
area, that is the piezometric line, is 
drawn as follows: cH = 195 feet is meas¬ 
ured horizontally on the base line of the 
pressure area, that is, at L. W. L. from a 















































I 



Fig. 94. View of Narora Weir and Headworka 












DAMS AND WEIRS 


167 



line through A to B. The point B is 
then joined with A on the head water 
level at the commencement of the rear 
apron. The hydraulic gradient will 
thus be 1 in 15. The intersection of 
this line BA with a vertical drawn 

through the first line of curtain is the 

; 

location of a step of two feet equal to 
the head absorbed in the vertical travel 
g >> at this point. Another line parallel to 

^ O 

Q | the hydraulic gradient is now drawn 
g | to the termination of the fore apron, 

3 GG 

Is "g this completes the piezometric line or 
§ s 2 the upper outline of the pressure area. 

* J > With regard to the floor thickness 
2 g at the toe of the drop wall, the value 

^ Ph '3 A 

* .2 £• of h, or loss of head due to percola- 
| g tion under the rear apron, is 6 feet, 

§ g from the rear curtain, 4 feet; total 10 
“I feet. H — h is, therefore, 13 — 10 = 3. 

^ a 3 3 

J o The thickness of the floor according 
^ .g § to formula (33) where (H — h) =3 feet 
3 * a comes to |X3 = 4 feet, the value of p 

* § * being assumed at 2. The floor natUr- 
-1 ally tapers toward its end where the 

5 | £ uplift is nil. The thickness at this 

* .2 3 point is made 3 feet which is about the 

® minimum limit. There remains now 
® I the talus of riprap, its length from 

bfi 

s £ formula (35) is 

L = 10 c ^”^j\ yv = 150 feet = 10c 

The thickness of the talus is generally 
four—often five feet—and is a matter 
of judgment considering the nature of 
the material used. 

125. Discussion of Narora Weir. 
The Narora weir itself, Fig. 94, forms 





















































168 


DAMS AND WEIRS 


a most instructive object lesson, demonstrating what is the least 
correct base width, or length of percolation consistent with absolute 
safety, that can be adopted for sands of class 2. The system of 
analyzing graphically an existing work with regard to hydraulic 
gradient is exemplified in Fig. 95 under three separate conditions; 
first, as the work originally stood, with a hydraulic gradient of 1 in 
11.8; second, at the time of failure, when the floor and the grouted 
riprap blew up. On this occasion owing to the rear apron having 
been washed away by a flood the hydraulic grade fell to 1 in 8; 
third, after the extension of the rear apron and curtailment of the 
fore apron had been effected. Under the first conditions the hori¬ 
zontal component of the length of travel or percolation L from A 
to E is 123 feet. The total length is made up of three parts: First, 
a step down and up in the foundation of the drop wall of 7 feet; 
second, a drop down and up of 12 feet either side of the downstream 
curtain wall; third, the horizontal distance 123 as above. The rise 
at the end of the floor is neglected. The total value of L is then 
123+7+12 + 12 = 154. This is set out on a horizontal line to the 
point C. AC is then the hydraulic gradient. 

This demonstrates that the hydraulic gradient was originally 
something under 1 in 12, and in addition to this the floor is very 
deficient in thickness. The hydrostatic pressure on the floor at the 
toe of the drop wall is 8 feet. To meet this the floor has a value of 
tp of only 5 feet. The specific gravity of the floor will not exceed 
2, as it was mostly formed of broken brick concrete in hydraulic 
mortar. The value of p — 1 will, therefore, be unity, the floor being 
submerged. In spite of this, the work stood intact to all external 
appearance for twenty years, when a heavy freshet in the river set 
up a cross current which washed out that portion of the rear apron 
nearest the drop wall, thus rendering the rest useless, the connection 
having been severed. 

On this occurrence, failure at once took place, as the floor had 
doubtless been on the point of yielding for some time. In fact, this 
state of affairs had been suspected, as holes bored in the floor very 
shortly before the actual catastrophe took place showed that a large 
space existed below it, full, not of sand, but of water. Thus the 
floor was actually held up by the hydrostatic pressure; otherwise it 
must have collapsed. The removal of the rear apron caused this 


DAMS AND WEIRS 


169 


pressure to be so much increased that the whole floor, together with 
the grouted pitching below the curtain, blew up. 

The hydraulic gradient BC is that at the time of the collapse. 
It will be seen that it is now reduced to 1 in 8. The piezometric 
line is not shown on the diagram. 

In restoring the work the rear apron was extended upstream as 
shown dotted in Fig. 95, to a distance of 80 feet beyond the drop wall, 
and was made five feet thick. It was composed of puddle covered 
with riprap and at its junction with the drop wall was provided 
with a solid masonry covering. The puddle foundation also was 
sloped down to the level of the floor base to form a ground connec¬ 
tion with the drop wall. At its upstream termination sheet piling 
was driven to a depth of twelve feet below floor level. 

The grouted pitching in the fore apron was relaid dry, except 
for the first ten feet which was rebuilt in mortar, to form a continua¬ 
tion of the impervious floor. Omitting the mortar has the effect 
of reducing the pressure on the floor. Even then the uplift would 
have been too great, so a water cushion 2 feet deep was formed over 
the floor by building a dwarf wall of concrete (shown on the section) 
right along its edge. This adds 1 foot to the effective value of tp. 
It will be seen that the hydraulic gradient now works out to 1 in 15. 
A value for c of 15 has been adopted for similar light sands from 
which that of other sands, as Classes I and III , have been deduced. 

It will be noticed that the crest of this weir is furnished with 
shutters which are collapsible when overtopped and are raised by 
hand or by a traveling crab that moves along the crest, raising the 
shutters as it proceeds. The shutters are 3 feet deep and some 
20 feet long. They are held up against the water by tension rods 
hinged to the w T eir, and at about J the height of the shutter, i.e., 
at the center of pressure. 

126. Sloping Apron Weirs, Type B. Another type of weir, 
designated B, will now be discussed, in which there is no direct 
vertical drop, the fore apron not being horizontal but sloping from 
the crest to the L. W. L. or to a little above it, the talus beyond 
being also on a flat slope or horizontal. 

In the modern examples of this type which will be examined, 
the height of the permanent masonry weir wall is greatly reduced, 
with the object of offering as little obstruction as possible to the 


170 


DAMS AND WEIRS 


passage of flood water. The canal level is maintained by means of 
deep crest shutters. In the Khanki weir, Fig. 96, the weir proper, 
or rather bar wall, is 7 feet high above L. W. L., while the shutters 
are 6 feet high. It, therefore, holds up 13 feet of water, the same 
as was the case with the Narora weir. 

The object of adopting the sloping apron is to avoid construc¬ 
tion in wet foundations, as most of it can be built quite in the dry 
above L. W. L. The disadvantage of this type lies in the con¬ 
striction of the waterway below the breast wall, which causes the 
velocity of overfall to be continued well past the crest. With a 
direct overfall, on the other hand, a depth of 7 feet for water to 
churn in would be available at this point. This would check the 
flow and the increased area of the waterway rendered available 



Fig. 96. Profile of Khanki Weir, Showing Restoration Work Similar to that of 

Narora Weir 


should reduce the velocity. For this reason, although the action 
on the apron is possibly less, that on the talus and river bed beyond 
must be greater than in the drop wall of type A. 

This work, like the former, failed for want of sufficient effective 
base length, and it consequently forms a valuable object lesson. 

As originally designed, no rear apron whatever, excepting a 
small heap of stone behind the breast wall, was provided. The 
value of L up to the termination of the grouted pitching is but 108 
feet; whereas it should have been cH or 15X13 = 195 feet. The 
hydraulic gradient, as shown in Fig. 96b, is only 1 in 8.3. This 
neglects the small vertical component at the breast wall. In spite 
of this deficiency in effective base width, the floor, owing to good 
workmanship, did not give way for some years, until gradually 
increased piping beneath the base caused its collapse. 



















































DAMS AND WEIRS 


171 


Owing to the raised position of the apron, it is not subject to 
high hydrostatic pressure. At its commencement it is ten feet 
below the summit level and nine feet of water acts at this point. 
r Ihis is met by four feet of masonry unsubmerged, of s.g. 2, which 
almost balances it. Thus the apron did not blow up, as was the case 
with the Narora weir, but collapsed. 

Some explanation of the graphical pressure diagram is required, 
as it offers some peculiarities, differing from the last examples. 
The full head, or II, is 13 feet. Owing, however, to the raised and 
sloping position of the apron, the base line of the pressure area will 
not be horizontal and so coincide with the L. W. L., but will be an 
inclined line from the commencement a to the point b, where the 
sloping base coincides with the L. W. L. From b where L. W. L. is 
reached onward, the base will be horizontal. With a sloping apron 
the pressure is nearly uniform, the water-pressure area is not wedge 
shaped but approximates to a rectangle. The apron, therefore, is 
also properly rectangular in profile, whereas in the overfall type 
the profile is, or should be, that of a truncated wedge. 

127. Restoration of Khanki Weir. After the failure of this 
work the restoration was on verv similar lines to that of the Narora 
weir. An impervious rear apron, seventy feet long, was constructed 
of puddle covered with concrete slabs, grouted in the joints. A rear 
curtain wall consisting of a line of rectangular undersunk blocks 
twenty feet deep, was provided. These additions have the effect 
of reducing the gradient to 1 in 16. The masonry curtain having 
regard to its great cost is of doubtful utility. A further prolonga¬ 
tion of the rear apron or else a line of sheet piling would, it is deemed, 
have been equally effective. Reinforced-eoncrete sheet piling is 
very suitable for curtain walls in sand and is bound to supplant the 
ponderous and expensive block curtain walls which form so marked 
a feature in Indian works. 

128. Merala Weir. Another weir on the same principle and 
quite recently constructed is the Merala weir at the head of the 
same historic river, the Chenab, known as the “Hydaspes” at the 
time of Alexander the Great. 

This weir, a section of which is given in Fig. 97, is located in the 
upper reaches of the river and is subjected to very violent floods; 
consequently its construction has to be abnormally strong to resist 


172 


DAMS AND WEIRS 



• H 

o 


£ 


d 


li 

a> 


‘VI 

o 

a 

o 


o 

a> 


m 


03 

s 


the dynamic action of the water. This 
is entirely a matter of judgment and 
no definite rules can possibly be given 
which would apply to different condi¬ 
tions. From a hydrostatic point of 
view the two lower lines of curtain 
blocks are decidedly detrimental and 
could well be cut out. If this were 
done the horizontal length of travel 
or percolation will come to 140. The 
head is 12 or 13 feet. If the latter, c 
having the value 15 as in the Khanki 
weir, the value of L will be 15X13 = 195 
feet. The horizontal length of travel 
is 140 feet and the wanting 55 feet 
will be just made up by the rear cur¬ 
tain. The superfluity of the two fore 
lines is thus apparent with regard to 
hydrostatic requirements. The long 
impervious sloping apron is a necessity 
to prevent erosion. 

It is a question whether a line of 
steel interlocking sheet piling is equally 
efficient as a curtain formed of wells 
of brickwork 12X8 feet undersunk and 
connected with piling and concrete 
filling. The latter has the advantage 
of solidity and weight lacking in the 
former. The system of curtain walls 
of undersunk blocks is peculiar to 
India. In the Hindia Barrage, in 
Mesopotamia, Fig. 115, interlocking 
sheet piling has been largely employed 
in places where well foundations would 
have been used in India. This change 
is probably due to the want of skilled 
well sinkers, who in India are extremely 
expert and form a special caste. 














































































DAMS AND WEIRS 


173 


The rear apron, in the Merala weir is of as solid construction 
as the fore apron and is built on a slope right up to crest level; this 
arrangement facilitates discharge. The velocity of approach must 
be very great to necessitate huge book blocks of concrete 6X6X3 
feet being laid behind the slope and beyond that a 40-foot length 
of riprap. The fore apron extends for 93 feet beyond the crest, 
twice as long as would be necessary with a weir of type A under 
normal conditions. The distance L of the talus is 203 feet against 
182 feet calculated from formula (35a). That of the lower weir at 
Khanki is 170 feet. This shows that the empirical formula gives 
a fair approximation. 

The fore apron in type B will extend to the toe of the slope or 
glacis. It is quite evident that the erosive action on a sloping apron 
of type B is far greater than that on the horizontal floor of type A; the 


Summit Level /? B 



uplift however is less, consequently the sloping apron can be made 
thinner and the saving thus effected put into additional length. 

129. Porous Fore ApronSo The next type of weir to be dealt 
with is type C. As it involves some fresh points, an investigation 
of it and the principles involved will be necessary. The previous 
examples of types A and B have been cases where the weir has as 
appendage an impervious fore apron which is subject to hydro¬ 
static pressure. There is another very common type which will be 
termed C, in which there is no impervious apron and the material 
which composes the body of the weir is not solid masonry but a 
porous mass of loose stone the only impervious parts being narrow 
vertical walls. In spite of this apparent contrariety it will be 
found that the same principle, viz, that of length of enforced per¬ 
colation, influences the design in this type as in the others. 











































174 


DAMS AND WEIRS 


Fig. 98 represents a wall upholding water to its crest and resting 
on a pervious substratum, as sand, gravel, or boulders, or a mixture 
of all three materials. The hydraulic gradient is AD; the upward 
pressure area ACD, and the base CD is the travel of the percolation. 
Unless this base length is equal to AC multiplied by the percolation 
factor obtained by experiment, piping will set in and the wall be 
undermined. Now as shown in Fig. 99, let a mass of loose stone 
be deposited below the wall. The weight of this stone will evi¬ 
dently have an appreciable effect in checking the disintegration and 
removal of the sand foundation. The water will not have a free 
untrammeled egress at D; it will, on the contrary, be compelled to 
rise in the interstices of the mass to a certain height EE determined 
by the extent to which the loose stones cause obstruction to the flow. 


Summit Level /I B 



Fig. 99. Effect on Percolation Due to Stones below Weir Wall of Fig. 98 

The resulting hydraulic gradient will now be A E —flatter than AD, 
but still too steep for permanency. 

In Fig. 100, the wall is shown backed by a rear apron of loose 
stone, and the fore apron extended to F. The water has now to 
filter through the rear apron underneath the wall and up through 
the stone filling in the fore apron. During this process a certain 
amount of sand will be washed up into the porous body and the 
loose stone will sink until the combined stone and sand forms a 
compact mass, offering a greater obstruction to the passage of the 
percolating water than exists in the sand itself and possessing far 
greater resisting power to disintegration. This will cause the level 
of water at E to rise until equilibrium results. When this is the 
case the hydraulic gradient is flattened to some point near F. If a 
sufficiently long body is provided, the resulting gradient will be 
equal to that found by experiment to produce permanent equilibrium. 














































DAMS AND WEIRS 


175 


The mass after the sinking process has been 
finished is then made good up to the original pro¬ 
file by fresh rock filling. At F near the toe of 
the slope the stone offers but little resistance 
either by its weight or depth; so it is evident that 
the slope of the prism should be flatter than the 
hydraulic gradient. 

The same action takes place with the rear 
apron, which soon becomes so filled with silt, as 
to be impervious or nearly so to the passage of 
water. But unless silt is deposited in the river 
bed behind, as eventually occurs right up to crest 
level, the thin portion of the rear slope, as well as 
the similar portion of the fore slope, cannot be 
counted as effective. Consequently out of the 
whole base length this part GF, roughly, about 
one-quarter, can be deemed inefficient as regards 
length of enforced percolation. As the consoli¬ 
dated lower part of the body of the weir gains 
in consistency, it can well be subject to hydro¬ 
static pressure. Consequently, the value of tp of 
the mass should be in excess of that of II — h, just 
as was the case with an impervious floor. 

130. Porous Fore Aprons Divided by Core 
Walls. In Fig. 101 a still further development is 
effected by the introduction of vertical body or 
core walls of masonry in the pervious mass of the 
fore apron. These impervious obstructions mate¬ 
rially assist the stability of the foundation, so 
much so that the process of underscour and set¬ 
tlement which must precede the balancing of the 
opposing forces in the purely loose stone mass 
need not occur at all, or to nothing like the same 
extent. If the party w'alls are properly spaced, 
the surface slope can be that of the hydraulic 
gradient itself and thus ensure equilibrium. This 
is clearly illustrated in the diagram. The water 
passing underneath the wall base CD will rise to 



Fig. 100. Effect on Percolation When Weir Wall Is Provided with Rear Apron of Loose Stone and Extensive Fore Apron of Stone Filling 





























Fig. 103. Madaya Rock-Fill Weir 



















































































































































DAMS AND WEIRS 


177 


the level F, the point E being somewhat higher; similar percolation 
under the other walls in the substratum will fill all the partitions full 
of water. The head AC will, therefore, be split up into four steps. 

Value of Rear Apron Very Great. The value of water tightness 
in the rear apron is so marked that it should be rendered impervious 
by a thick under layer of clay, and not left entirely to more or less 
imperfect surface silt stanching, except possibly in the case of high 
dams where a still settling pool is formed in rear of the work. 

131. Okhla and Madaya Weirs. In Fig. 102 is shown a 
detailed section of the Okhla rock-filled weir over the Jumna 
River, India. It is remarkable as being the first rock-filled weir 
not provided with any lines of curtain walls projecting below 
the base line, which has hitherto invariably been adopted. The 
stability of its sand foundation is consequently entirely depend¬ 
ent on its weight and its effective base length. As will be 
seen, the section is provided with two body walls in addition to 
the breast wall. The slope of the fore apron is 1 in 20. It is 
believed that a slope of 1 in 15 would be equally effective, a hori¬ 
zontal talus making up the continuation, as has been done in the 
Madaya weir, Fig. 103, which is a similar work but under much 
greater stress. 

The head of the water in the Okhla weir is 13 feet, with shutters 
up and weir body empty of water—a condition that could hardly 
exist. This would require an effective base length, L, of 195 feet; 
the actual is 250 feet. But, as noted previously, the end parts of 
the slopes cannot be included as effective; consequently the hydraulic 
gradient will not be far from 1 in 15. The weight of the stone, or 
tp, exactly balances this head at the beginning, as it is 10X1.3 = 13 
feet. If the water were at crest level and the weir full of water, 
tp would equal 8 feet, or rather a trifle less, owing to the lower level 
of the crest of the body wall. The head of 13 feet is broken up 
into four steps. The first is 3 feet deep, acting on a part of the 
rear apron together with 30 feet of the fore apron, say, 1 in 15; the 
rest are 1 in 20. A slope of 1 in 15 for the first party wall would 
cut the base at a point 40 feet short of the toe. Theoretically a 
fourth party wall is required at this point, but practically the rip¬ 
rap below the third dwarf wall is so stanched with sand as not to 
afford a free egress for the percolation; consequently the slope may 


DAMS AND WEIRS 



be assumed to continue on to its inter¬ 
section with the horizontal base. As 
already noted, material w T ould be saved 
in the section by adopting a reliably 
stanch rear apron and reducing the fore 
slope to 1 in 15, with a horizontal con¬ 
tinuation as was done in the Madaya 
weir. 

Economy of Type C. This type C 
is only economical where stone is abun¬ 
dant. It requires little labor or masonry 
work. On the other hand, the mass of 
the material used is very great, much 
greater, in fact, than is shown by the sec¬ 
tion. This is owing to the constant sink¬ 
ing and renewal of the talus which goes 
on for many years after the first construc¬ 
tion of the weir. 

The action of the flood on the talus 
is undoubtedly accentuated by the con¬ 
traction of the waterway due to the high 
sloping apron. The flood velocity 20 feet 
below the crest has been gaged as high 
as 18 feet per second. This would be 
very materially reduced if the A type of 
overfall were adopted, as the area of 
waterway at this point would be more 
than doubled. 

132. Dehri Weir. Another typical 
example of this class is the Dehri weir 
over the Son River, Fig. 104. The value 
of L , if the apexes of the two triangles of 
stone filling are deducted and the cur¬ 
tain walls included, comes to about 12 H f 
12 being adopted for this class of coarse 
sand. The curtain walls, each over 12,500 
feet long, must have been enormously 
costly. From the experience of Okhla, 




























































DAMS AND WEIRS 


179 


a contemporary work, on a much worse class 
of sand, curtain protection is quite unneces¬ 
sary if sufficient horizontal base width is 
provided. The head on this weir is 10 feet, 
and the height of breast wall 8 feet, tp is, 
therefore, 1.3X8=10.1, which is sufficient, 
considering that the full head will not act 
here. The lines of curtains could be safely 
dispensed with if the following alterations 
were made: (1) Rear apron to be reliably 
stanched in order to throw back the incidence 
of pressure and increase the effective base 
length; (2) three more body walls to be intro¬ 
duced; (3) slope 1 in 12 retained, but base to 
be dredged out toward apex to admit of no 
thickness under five feet. This probably 
would not cause any increase in the quantities 
of masonry above what they now are, and 
would entirely obviate the construction of 
nearly five miles of undersunk curtain blocks. 

133. Laguna Weir. The Laguna weir 
over the Colorado River, the only example of 
type C in the Lmited States is shown in Fig. 
105. Compared with other examples it might 
be considered as somewhat too wide if regard 
is had to its low unit flood discharge, but the 
inferior quality of the sand of this river 
probably renders this necessary. The body 
walls are undoubtedly not sufficiently numer¬ 
ous to be properly effective. The provision 
of an impervious rear apron would also be 
advantageous. 

134. Damietta and Rosetta Weirs. The 

location of the Damietta and Rosetta subsid¬ 
iary weirs, Fig. 106, which have been rather 
recently erected below the old Nile barrage, 
is shown in Fig. 92. These weirs are of type 
C, but the method of construction is quite 


o 


!o 


s 



Fig. 105. Section of Laguna Weir over Colorado River 










































180 


DAMS AND WEIRS 



novel and it is this alone that renders this 
work a valuable object lesson. The deep 
foundation of the breast wall was built 
without any pumping, all material having 
been deposited in the water of the Nile 
River. First the profile of the base was 
dredged out, as shown in the section. Then 
the core wall was constructed by first 
depositing, in a temporary box or enclosure 
secured by a few piles, loose stone from 
barges floated alongside. The whole was 
then grouted with cement grout, poured 
through pipes let into the mass. On the 
completion of one section all the appliances 
were moved forward and another section 
built, and so on until the whole wall was 
completed. Clay was deposited at the base 
of the core wall and the profile then made 
up by loose stone filling. 

This novel system of subaqueous con¬ 
struction has proved so satisfactory that in 
many cases it is bound to supersede older 
methods. Notwithstanding these innova¬ 
tions in methods of rapid construction, the 
profile of the weir itself is open to the objec¬ 
tion of being extravagantly bulky even for 
the type adopted, the base having been 
dredged out so deep as to greatly increase 
the mean depth of the stone filling. 

It is open to question whether a row 
or two rows of concrete sheet piles would 
not have been just as efficacious as the deep 
breast wall, and would certainly have been 
much less costly. The pure cement grout¬ 
ing was naturally expensive, but the admix¬ 
ture of sand proved unsatisfactory as the 
two materials of different specific gravity 
separated and formed layers; consequently, 



























DAMS AND WEIRS 


181 


pure cement had to be used. It may be noted that the value of L 
here is much less than would be expected. At Narora weir it is 11c, 
or 165 feet. Here, with a value of c of 18, it is but 8 \c, or 150 feet, 
instead of 200 feet, according to the formula. This is due to the 
low flood velocity of the Nile River compared with the Ganges. 

135. The Paradox of a Pervious Dam. From the conditions 
prevailing in type C it is clear that an impervious apron as used in 
types A and B is not absolutely essential in order to secure a safe 
length of travel for the percolating subcurrent. If the water is 
free to rise through the riprap and at the same time the sand in the 
river bed is prevented from rising with it, the practical effect is the 
same as with an impervious apron. “Fountaining”, as spouting 
sand is technically termed, is prevented and consequently also 
“piping”. This latter term defines the gradual removal of sand 
from beneath a foundation by the action of the percolating under 
current. Thus the apparent paradox that a length of filter bed, 
although pervious, is as effective as a masonry apron would be. 
The hydraulic gradient in such case will be steeper than allowable 
under the latter circumstance. Filter beds are usually composed 
of a thick layer of gravel and stone laid on the sand of the river 
bed, the small stuff at the bottom and the larger material at the top. 
The ideal type of filter is one composed of stone arranged in sizes as 
above stated of a depth of 4 or 5 feet covered with heavy slabs or 
book blocks of concrete; these are set with narrow open intervals 
between blocks as shown in Figs. 96 and 97. Protection is thus 
afforded not only against scour from above but also from uplift 
underneath. Although the subcurrent of water can escape through 
a filter its free exit is hindered, consequently some hydrostatic 
pressure must still exist below the base, how much it is a difficult 
matter to determine, and it will therefore be left out of considera¬ 
tion. If the filter bed is properly constructed its length should be 
included in that of L or the length of travel. Ordinary riprap, 
unless exceptionally deep, is not of much, if any, value in this 
respect . The Hindia Barrage in Mesopotamia, Fig. 115, section 145, is 
provided with a filter bed consisting of a thick layer of stone 65.5 feet 
wide which occurs in the middle of the floor. The object of this 
is to allow the escape of the subcurrent and reduce the uplift on 
the dam and on that part of the floor which is impervious. 


182 


DAMS AND WEIRS 


136. Crest Shutters. Nearly all submerged river weirs are 
provided with crest shutters 3 to 6 feet deep, 6 feet. being the 
height adopted in the more recent works. These are generally 
raised by means of a traveling crane running on rails just behind 
the hinge of the gate. When the shutters are tripped they fall 
over this railway. In the case of the Merala weir, Fig. 97, the 
raising of the shutters is effected from a trolley running on overhead 
wires strung over steel towers erected on each pier. These piers or 
groins are 500 feet apart. The 6-foot shutters are 3 feet wide, 
held up by hinged struts which catch on to a bolt and are easily 
released by hand or by chains worked from the piers. On the 
Betwa weir the shutters, also 6 feet deep, are automatic in action, 
being hinged to a tension rod at about the center of pressure, con¬ 
sequently when overtopped they turn over and fall. Not all are 
hinged at the same height; they should not fall simultaneously but 
ease the flood gradually. The advantage of deep shutters is very 
great as the permanent weir can be built much lower than other¬ 
wise would be necessary, and thus offer much less obstruction to 
the flood. The only drawback is that crest shutters require a resi¬ 
dent staff of experienced men to deal with them. 

The Laguna weir, Fig. 105, has no shutters. The unit flood 

discharge of the Colorado is, however, small compared with that of 

the Indian rivers, being only 22 second-feet, whereas the Merala 

weir discharges 150 second-feet per foot run of weir, consequently 

shutters in the former case are unnecessarv. 

*/ 

OPEN DAMS OR BARRAGES 

137. Barrage Defined. The term “open dam”, or barrage, 
generally designates what is in fact a regulating bridge built across 
a river channel, and furnished with gates which close the spans as 
required. They are partial regulators, the closure being only 
effected during low water. When the river is in flood, the gates 
are opened and free passage is afforded for flood water to pass, the 
floor being level with the river bed. Weir scouring sluices, which 
are indispensable adjuncts to weirs built over sandy rivers, belong 
practically to the same category as open dams, as they are also 
partial regulators, the difference being that they span only a por¬ 
tion of the river instead of the whole, and further are subject to great 


DAMS AND WEIRS 


183 


scouring action from the fact that when the river water is artificially 
raised above its normal level by the weir, the downstream channel 
is empty or nearly so. 

function of Weir Sluices. The function of weir sluices is two¬ 
fold: First, to train the deep channel of the river, the natural 
course of which is obliterated by the weir, past the canal head, and 
to retain it in this position. Otherwise, in a wide river the low 
water channel might take a course parallel to the weir crest itself, 
or else one distant from the canal head, and thus cause the approach 
channel to become blocked with deposit. 

Second, by manipulating the sluice gates, silt is allowed to 
deposit in the slack water in the deep channel. The canal is thereby 
preserved from silting up, and when the accumulation becomes 
excessive, it can be scoured out by opening the gates. 

The sill of the weir sluice is placed as low as can conveniently 
be managed, being generally either at L. W. L. itself, or somewhat 
higher, its level generally corresponding with the base of the drop 
or breast wall. Thus the maximum statical head to which the work 
is subjected is the height of the weir crest plus that of the weir 
shutters, or II 

The ventage provided is regulated by the low-water discharge 
of the river, and should be capable of taking more than the average 
dry season discharge. In one case, that of the Laguna weir, where 
the river low supply is deficient, the weir sluices are designed to take 
the whole ordinary discharge of the river excepting the highest 
floods. This is with the object of maintaining a wide, deep channel 
which may be drawn upon as a reservoir. This case is, however, 
exceptional. 

As the object of a weir sluice is to pass water at a high velocity 
in order to scour out deposit for some distance to the rear of the 
work, it is evident that the openings should be wide, with as few 
obstructions as possible in the way of piers, and should be open at 
the surface, the arches and platform being built clear of the flood 
level. Further, in order to take full advantage of the scouring 
power of the current, which is at a maximum at the sluice itself, 
diminishing in velocity with the distance to the rear of the work, 
it is absolutely necessary not only to place the canal head as close 
as possible to the weir sluices, but to recess the head as little 


184 


DAMS AND WEIRS 



ZO 

1 


Scale of Feet 

O ZO 40 60 , 

iii i i i 

Flevahons refer to 


60 80 /OO 
Sea Level 


Fig. 107. Plan of Laguna Weir-Scouring Sluices 



































DAMS AND WEIRS 


185 


as practicable behind the face line of the abutment of the end 
sluice vent. 

With regard to canal head regulators or intakes, the regulation 
effected by these is entire, not partial, so that these works are sub¬ 
jected to a much greater statical stress than weir sluices, and conse¬ 
quently, for convenience of manipulation, are usually designed with 
narrower openings than are necessary or desirable in the latter. 
The design of these works is, however, outside the scope of the 
subject in hand. 

138. Example of Weir Scouring Sluice. Fig. 107 is an excel¬ 
lent example of a weir scouring sluice, that attached to the Laguna 



Fig. 108. View of Yuma Canal and Sluiceway Showing Sluice 

Gates under Construction 


weir, the profile of which was given in Fig. 105. The Yuma 
canal intake is placed clear of the sandy bed of the river on a rock 
foundation and the sluiceway in front of it is also cut through solid 
rock independent of the weir. At the end of this sluiceway and just 
past the intake the weir sluices are located, consisting of three spans 
of 33| feet closed by steel counterweighted roller gates which can be 
hoisted clear of the flood by electrically operated winches. The 
gates are clearly shown in Fig. 108, which is from a photograph 
taken during the progress of the work. The bed of the sluiceway 
is at El. 138.0, that of the canal intake sill is 147.0, and that of the 












186 


DAMS AND WEIRS 


weir crest 151.0—hence the whole sluiceway can be allowed to fill 
up with deposit to a depth of 9 feet, without interfering with the 



discharge of the canal, or if the dashboards of the intake are lowered 
the sluiceway can be filled up to El 156 which is the level of the top 
of the draw gates, i.e., 18 feet deep. The difference between high 














































































DAMS AND WEIRS 


187 



Fig. 110. Design Diagrams for Weir Sluices of Corbett Dam 









































































































































































188 


DAMS AND WEIRS 



Fig. 111. View of Corbett Dam on Shoshone River in Winter 














DAMS AND WEIRS 


189 



water above and below the sluice gates 
is 11 feet, consequently when the gates 
are lifted immense scour must take place 
and any deposit be rapidly removed. 
The sluiceway is in fact a large silt 
trap. 

139. Weir Sluices of Corbett Dam. 

The weir sluices of the Corbett dam on 
the Shoshone River, Wyoming, are given 
in Figs. 109, 110, and 111. 

The canal takes out through a tun¬ 
nel, the head of which has necessarily to 
be recessed far behind the location of the 
weir sluices. Unless special measures 
were adopted, the space between the 
sluice gates and the tunnel head would 
fill up with sand and deposit and block 
the entrance. 

To obviate this a wall 8 feet high is 
built encircling the entrance. A “divide” 
wall is also run out upstream of the weir 
sluices, cutting them off from the weir 
and its approaches. The space between 
these two walls forms a sluiceway which 
draws the current of the river in a low 
stage past the canal head and further 
forms a large silt trap which can be 
scoured out when convenient. Onlv a 
thin film of surface water can overflow 
the long encircling wall, then it runs down 
a paved warped slope which leads it into 
the head gates, the heavy silt in suspen¬ 
sion being deposited in the sluiceway. 
This arrangement is admirable. 

The fault of the weir sluices as built 
is the narrowness of the openings which 
consist of three spans of 5 feet. One 
span of 12 feet would be much more 










































































190 


DAMS AND WEIRS 


effective. In modern Indian practice, weir sluices on large rivers 
are built with 20 to 40 feet openings. 

140. Weir Scouring Sluices on Sand. Weir scouring sluices 
built on pure sand on as large rivers as are met with in India are 
very formidable works, provided with long aprons and deep lines 
of curtain blocks. An example is given in Fig. 112 of the so-termed 
undersluices of the Khanki weir over the Chenab River in the Pun¬ 
jab. The spans are 20 feet, each closed by 3 draw gates, running 
in parallel grooves, fitted with antifriction wheels (not rollers), lifted 
by means of traveling power winches which straddle the openings 
in which the grooves and gates are located. 

The Merala weir sluices of the Upper Chenab canal have 8 
spans of 31 feet, piers feet thick, double draw gates 14 feet high. 



Fig. 113. View of Merala Weir Sluices, Upper Chenab Canal 


These are lifted clear of the flood, which is 21 feet above floor, by 
means of steel towers 20 feet high erected on each pier. These carry 
the lifting apparatus and heavy counterweights. These gates, like 
those at Laguna weir, Fig. 108, bear against Stoney roller frames. 

Fig. 113 is from a photograph of the Merala weir sluices. The 
work is a partial regulator, in that complete closure at high flood is 
not attempted. The Upper Chenab canal is the largest in the world 
with the sole exception of the Ibramiyah canal in Egypt, its dis¬ 
charge being 12,000 second-feet. Its depth is 13 feet. The capacity 
of the Ibramiyah was 20,000 second-feet prior to head regulation. 

141. Heavy Construction a Necessity. In works of this 
description solid construction is a necessity. Light reinforced con¬ 
crete construction would not answer, as weight is required, not only 















DAMS AND WEIRS 


191 


to withstand the hydrostatic pressure but the dynamic effects of 
flood water in violent motion. Besides which widely distributed 



weight is undoubtedly necessary for works built on the shifting 
sand of a river bed, although this is a matter for which no definite 
rules can be formed. 
































192 


DAMS AND WEIRS 


The weir sluices at Laguna and also at the Corbett dam, are 
solid concrete structures without reinforcement. 

In the East, generally, reinforced concrete is not employed nor 
is even cement concrete except in wet foundations, the reason 
being that cement, steel, and wood for forms are very expensive 
items whereas excellent natural hydraulic lime is generally avail¬ 
able, skilled and unskilled labor is also abundant. A skilled mason’s 
wages are about 10 to 16 cents and a laborer’s 6 to 8 cents for a 
12-hour day. L T nder such circumstances the employment of rein¬ 
forced cement concrete is entirely confined to siphons where tension 
has to be taken care of. 

In America, on the other hand, the labor conditions are such 
that reinforced concrete which requires only unskilled labor and is 
mostly made up by machinery, is by far the most suitable form of 
construction from point of view of cost as well as convenience. 

This accounts for the very different appearance of irrigation 
works in the East from those in the West. Both are suitable under 
the different conditions that severally exist. 

142. Large Open Dams across Rivers. Of open dams built 
across rivers, several specimens on a large scale exist in Egypt. 
These works, like weir sluices are partial regulators and allow free 
passage to flood water. 

Assiut Barrage. In the Assiut barrage, Figs. 114 and 115,* 
constructed across the Nile above the Ibramiyah canal head 
in lower Egypt, the foundations are of sand and silt of a worse 
quality than is met with in the great Himalayan rivers. The 
value of c adopted for the Nile is IS, against 15 for Himalayan rivers. 
This dam holds up 5 meters of water, the head or difference of 
levels being 3 meters. Having regard to uplift, the head is the 
difference of levels but when considering overturning moment, on 
IP h 3 

the piers, —-— is the moment, II and h being the respective 

6 6 

depths of water above and below the gates. It is believed that in 
the estimation of the length of travel the vertical sheet piling was 
left out of consideration. Inspection of the section in Fig. 115 


* In Figs. 115 and 116 and in the discussion of these problems in the text, the metric dimen¬ 
sions used in the plans of the works have been retained. Meters multiplied by the factor 
3.28 will give the proper values in feet. 




El. 57.00 


DAMS AND WEIRS 


193 



shows that the foundation is 
mass cement concrete, 10 feet 
deep, on which platform the 
superstructure is built. This 
latter consists of 122 spans of 5 
meters, or 16 feet, with piers 2 
meters thick, every ninth being 
an abutment pier 4 meters thick 
and longer than the rest. This 
is a work of excessive solidity 
the ratio of thickness of piers 
to the span being .48, a propor¬ 
tion of .33 S would, it is con¬ 
sidered be better. This could 
be had by increasing the spans 
to 6 meters, or 20 feet right 
through, retaining the pier thick¬ 
ness as it is at present. 

143. General Features of 
River Regulators. All these river 
regulators are built on the same 
general lines, viz, mass founda¬ 
tions of a great depth, an arched 
highway bridge, with spring of 
arch at flood level, then a gap 
left for insertion of the double 
grooves and gates, succeeded by 
a narrow strip of arch sufficient 
to carry one of the rails of the 
traveling winch, the other rest¬ 
ing on the one parapet of the 
bridge. 

The piers are given a bat¬ 
ter downstream in order to bet¬ 
ter distribute the pressure on 
the foundation. The resultant 
of the weight of one span com¬ 
bined with the horizontal water 






































































































































194 


DAMS AND WEIRS 


pressure must fall within the middle third of the base of the pier, 
the length of which can be manipulated to bring this about. In 
this case it does so even with increase of the span to 6 meters. 
This combined work is of value considered from a military point 
of view, as affording a crossing of the Nile River; consequently 
the extreme solidity of its construction was probably considered a 
necessity. 

In some regulators girders are substituted for arches, in others 
as we have noted with regard to the Merala weir sluices, the super¬ 
structure above the flood line is open steel work of considerable 
height. 


144. Stability of Assiut Barrage. The hydraulic gradient in 
Fig. 115, neglecting the vertical sheet piling, is drawn on the profile 
and is the line AB, the horizontal distance is 43 meters while the 
head is 3 meters. The slope is therefore 1 in 14J. The uplift is 
the area enclosed between AB and a horizontal through B which is 
only 1.4 meters at its deepest part near the gates. Upstream of 
the gates the uplift is more than balanced by the weight of water 
overlying the floor. The horizontal travel of the percolation is from 
A to B plus the length of the filter as explained in section 135. 

The horizontal travel is therefore 51§ meters and the ratio or 


. 51.5 
C,1S ~3“ 


17.2. The piezometric line has also been shown, includ¬ 


ing in this case the two vertical obstructions. Their effect on the 
uplift is very slight, owing to the fore curtain which raises the grade 
line. The slope in this case is obtained by adding the vertical to 
the horizontal travel, i.e., from B to D, BC and CD being 8 meters 
each in length, AD is then the hydraulic gradient which is 1 in 23. 
Steps occur at points b and c; for instance the line AB is part of 
AD, the line be is parallel to AD drawn up from C , and the line 
cB is similarly drawn up from B forming the end step. 

This work is the first to be built with a filter downstream, 
which has the practical effect of adding to the length of percolation 
travel irrespective of the hydraulic grade. 

145. The Hindia Barrage. The Hindia barrage, quite recently 
erected over the Euphrates River near Bagdad, is given in Fig. 116. 
This work, which was designed by Sir William Willocks, bears a 



DAMS AND WEIRS 


195 


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close resemblance to 
the Egyptian regu¬ 
lators, viz, the Assiut, 
the Zifta, and other 
works constructed 
across the river Nile. 
The piers are reduced 
to 1.50 meters from 
the 2-meters thick¬ 
ness in the Assiut 
dam, Fig. 115, and 
there are no abut¬ 
ment piers, conse¬ 
quently the elevation 
presents a much 
lighter appearance. 
The ratio of thick¬ 
ness to span is .3. 
In order to reduce 
the head on the work, 
a filter bed 20 meters 
wide is introduced 
just beyond the plat¬ 
form of the founda¬ 
tion of the regulat¬ 
ing bridge. The 
upward pressure is 
thus presumed to be 
nil at the point D. 
The head is the dis¬ 
tance between the 
summer supply level 
upstream, and that 
downstream above 
the subsidiary weir, 
this amounts to 3.50 
meters. The length 
of compulsory travel 
















































































































































196 


DAMS AND WEIRS 


from A to B including .50 meter due to the sheet piling is 36.50 

36.5 

meters. AB is then the hydraulic gradient, which is 1 in -r-jr 

o.o 

= 1 in 10.4. The piezometric line DFC is drawn up from D parallel 
to AB. The area of uplift is DGHEF; that part of the uplift 
below the line DE is however accounted for by assuming all 
masonry situated below El. 27.50 as reduced in weight by flota¬ 
tion, leaving the area DEF as representing the uplift still unac¬ 
counted for. 

Beyond the filter is a 21-meter length of impervious apron con¬ 
sisting of clay puddle covered by stone paving, which abuts on a 
masonry subsidiary weir. This wall holds the water up one meter 
in depth and so reduces the head to that extent, with the further 
addition of the depth of film passing over the crest at low water 
which is .5 meter, total reduction 1.50 meters. 

This is the first instance of the use of puddle in a fore apron, 
or talus; its object is, by the introduction of an impervious rear 
apron 21 meters long, to prevent the subsidiary weir wall from being 
undermined. The head being 1| meters, the length of travel 
required, taking c as 18, will be 18X1.5 = 27 meters. The actual 
length of travel provided is vertical 15, horizontal 41, total 56 
meters, more than double what is strictly requisite. The long 
hearth of solid masonry which is located below the subsidiary drop 
wall is for the purpose of withstanding scour caused by the over- 
fall. Beyond this is the talus of riprap 20 meters wide and a row 
of sheet piling. The total length of the floor of this w T ork is 364 
feet, with three rows of sheet piles. That of the Assiut barrage 
is 216 feet with two rows of sheeting. The difference in head is 
half a meter only, so that certain unknown conditions of flood or 
that of the material in the bed must exist to account for the excess. 

146. American vs. Indian Treatment. In American regulat¬ 
ing works it is generally the fashion where entire closure is required 
to locate the draw gates and their grooves inside the panel or bulk¬ 
head wall that closes the upper part of the regulator above the 
sluice openings. Thus when the gates are raised they are concealed 
behind the panel walls. In Indian practice the gate grooves in the 
piers are generally located outside the bulkhead wall; thus when 
hoisted, the gates are visible and accessible. Fig. 117 is from a 



DAMS AND WEIRS 


197 


photograph of a branch head, illustrating this. The work is of 
reinforced concrete as can be told from the thinness of the piers. 
In an Indian work of similar character the pier noses would project 
well beyond the face wall of the regulator and the gates would be 
raised in front, not behind it. 

The use of double gates is universal in Eastern irrigation works; 
they have the following unquestionable advantages over a single 
gate: First, less power for each is required to lift two gates than 
one; second, when hoisted they can be stacked side by side and so 
the pier can be reduced in height; third, where sand or silt is in 
suspension, surface water can be tapped by leaving the lower leaf 
down while the upper is raised; and fourth, regulation is made easier. 



Fig. 117. Typical American Regulating Sluices in Reinforced 

Concrete Weir 


In the Khanki weir sluices, Fig. 112, 3 gates running in 3 grooves 
are employed. 

147. Length of Spans. In designing open dams the spans 
should be made as large as convenient, the tendency in modern 
design is to increase the spans to 30 feet or more; the Laguna weir 
sluices are 33J feet wide and the Merala 31 feet. The thickness of 
the piers is a matter of judgment and is best expressed as some 
function of the span, the depth of water by which the height of the 
piers is regulated, forms another factor. 

The depth of water upheld regulates the thickness more than 
the length. The length should be so adjusted that the resultant 
line of pressure combined of the weight of one pier and arch, or 
superstructure and of the water pressure acting on one span falls 
within the middle third of the base. 






198 


DAMS AND WEIRS 


For example take the Assiut regulator, Fig. 115. The con¬ 
tents of one pier and span allowing for uplift is roughly 390 cubic 
meters of masonry, an equivalent to 1000 tons. The incidence of 
W is about 2 meters from the middle third downstream boundary. 

The moment of the weight about this point is therefore 1000 X 
2=2000 meter tons. Let H be depth of water upstream, and h 

(Ii ^ — h?') wl 

downstream, then the overturning moment is expressed by--- 

Here H = 5,h = 2 meters, w = 1.1 tons per cubic meter, the length l 


of one span is 7 meters; then the moment = 


(125 —8)X1.1X7 


6 


= 150 


meter tons. The moment of resistance is therefore immensely in excess 
of the moment of water pressure. The height of the pier is however 
governed by the high flood level, the width by the necessity of a 
highway bridge. At full flood nearly the whole of the pier will be 
immersed in water and so lose weight. There is probably some 
intermediate stage when the water pressure will be greater than 
that estimated, as would be the case if the gates were left closed 
while the water topped them by several feet, the water downstream 
not having had time to rise to correspond. 

148. Moments for Hindia Barrage. In the case of the Hindia 
barrage, Fig. 116, H = 5 meters, h = 1.5, then 


M __ (125 —3.4) XLIX6.50 

6 


= 145 meter tons 


The weight of one span is estimated at 180 tons. Its moment 
about the toe of the base is about 180X6.5 = 1170 meter tons. 


The factor of safety against overturning is therefore 


1170 

145 


= 8 . 


The long base of these piers is required for the purpose of 
distributing the load over as wide an area as possible in order to 
reduce the unit pressure to about one long ton per square foot. 
This is also partly the object of the deep mass foundation. The 
same result could doubtless be attained with much less material by 
adopting a thin floor say two or three feet thick, reinforced by steel 
rods so as to ensure the distribution of the weight of the super¬ 
structure evenly over the whole base. It seems to the writer that 
the Assiut barrage with its mass foundation having been a success 






DAMS AND WEIRS 


199 



Fig. 118. Head Regulator and Undersluices of North Mon Canal in Burma, Showing Portion of Weir 






























200 


DAMS AND WEIRS 


TABLE II 


Pier Thickness—Suitable for Open Partial Regulators and Weir Sluices 



DEPTHS OF WATER 

SPAN 

15 Feet 

20 Feet 

25 Feet 

30 Feet 


M. 

T. 

M. 

T. 

M. 

T. 

M. 

T. 

10 feet 

.25 

2.5 

.27 

2.7 

.29 

2.9 

.31 

3.1 

15 feet 

.24 

3.6 

.26 

3.9 

.28 

4.2 

.30 

4.5 

20 feet 

.23 

4.6 

.25 

5.0 

.27 

5.4 

.29 

5.8 

25 feet 

.21 

5.3 

.24 

6.0 

.26 

6.5 

.28 

7.0 


M is multiplier of span for thickness T. 


as regards stability, is no reason why a heavy style of construction 
such as this should be perpetuated. 

149. North Mon Canal. In Fig. 118 is shown the head works 
of the Mon right canal in Burma. The weir is of type A, with crest 
shutters and sluices of large span controlled by draw gates. In the 
canal head, the gates are recessed behind the face wall as in American 
practice. 

150. Thickness of Piers. Table II, though purely empirical 
will form a useful guide of thickness of piers in open dams or partial 
regulators. 

If reinforced, very considerable reduction can be made in the 
thickness of piers, say §, but for this class of river work a heavy 
structure is obligatory. 

151. Advantages of Open Dams. Open dams have the follow¬ 
ing advantages over solid weirs, or combinations of solid and over- 
fall dams: First, the river bed is not interfered with and conse¬ 
quently the heading up and scour is only that due to the obstruction 
of the piers, which is inconsiderable. This points to the value of 
wide spans. Second, the “river low” supply is under complete 
control. Third, a highway bridge across the river always forms 
part of the structure which in most countries is a valuable asset. 

Open dams, on the other hand, are not suitable for torrential 
rivers as the Himalayan rivers near their points of debouchure 
from the mountains, or wherever such detritus as trees, logs, etc., 
are carried down in flood time. 






























DAMS AND WEIRS 


201 


152. Upper Coleroon Regulator. Fig. 119 is from a photo¬ 
graph of a regulating bridge on the upper Coleroon River in the 
Madras 1 residency, southern India. Originally a weir of type A was 
constructed at this site in conjunction with a bridge. The constric¬ 
tion of the discharge due to the drop wall, which was six feet high, 
and the piers of the bridge, caused a very high afflux and great scour 
on the talus. Eventually the drop wall was cleared away altogether, 
the bridge piers were lengthened upstream and fitted with grooves 
and steel towers, and counterweighted draw gates some 7 feet deep 



Fig. 119. View of Regulating Bridge on the Upper Coleroon River. Southern India 


took the place of the drop wall. In the flood season the gates can 
be raised up to the level of the bridge parapet quite clear of the 
flood. The work was thus changed from one of a weir of type A, 
to an open dam. The original weir and bridge were constructed 
about half a century ago. 

153. St. Andrew’s Rapids Dam. Another class of semi-open 
dam consists of a permanent low floor or dwarf weir built across 
the river bed which is generally of rock, and the temporary dam¬ 
ming up of the water is effected by movable hinged standards being 
lowered from the deck of an overbridge, which standards support 













202 


DAMS AND WEIRS 



Fig. 120. Elevations and Part Section of St. Andrew’s Rapids Semi-Open Dam 

















































































































































































































































































































































































































































DAMS AND WEIRS 


203 


either a rolled reticulated curtain let down to cover them or else a 
steel sliding shutter mounted on rollers. 

The St. Andrew’s Rapids dam, Fig. 120, a quite recent construc¬ 
tion, may be cited as an example. The object of the dam is to raise 
the water in the Red River, Manitoba, to enable steamboats to navi¬ 
gate the river from Winnipeg City to the lake of that name. To 
effect this the water level at the rapids has to be raised 20 feet 
above L. W. L. and at the same time, on account of the accumu¬ 
lation of ice brought down by the river, a clear passage is a necessity. 
The Red River rises in the South, in the State of North Dakota 
where the thaw sets in much earlier than at Lake Winnipeg, con¬ 
sequently freshets bring down masses of ice when the river and lake 
are both frozen. 

Camere Type of Dam. The dam is of the type known as the 
Camere curtain dam, the closure being effected by a reticulated 
wooden curtain, which is rolled up and down the vertical frames 
thereby opening or closing the vents. It is a French invention, 
having been first constructed on the Seine. The principle of this 
movable dam consists in a large span girder bridge, from which 
vertical hinged supports carrying the curtain frames are let drop 
on to a low weir. When not required for use these vertical girders 
are hauled up into a horizontal position below the girder bridge and 
fastened there. In fact, the principle is very much like that of a 
needle dam. The river is 800 feet wide, and the bridge is of six 
spans of 138 feet. 

The bridge is composed of three trusses, two of which are free 
from internal cross-bracing, and carry tram lines with all the work¬ 
ing apparatus of several sets of winches and hoists for manipulating 
the vertical girders and the curtain; the third truss is mainly to 
strengthen the bridge laterally, and to carry the hinged ends of the 
vertical girders. 

It will be understood that the surface exposed to wind pressure 
is exceptionally great, so that the cross-bracing is absolutely essen¬ 
tial, as is also the lateral support afforded by a heavy projection of 
the pier itself above floor level. 

In the cross-section it will be seen that there is a footbridge 
opening in the pier. This footbridge will carry winches for wind¬ 
ing and unwinding the curtains, and is formed by projections thrown 


204 


DAMS AND WEIRS 


out at the rear of each group of frames. It will afford through 
communication by a tramway. The curtains can be detached 
altogether from the frames and housed in a chamber in the pier 
clear of the floodline. 

The lower part of the work consists of a submerged weir of 
solid construction which runs right across the river; its crest is 7 
feet 6 inches above L. W. L. at El. 689.50. The top of the curtains 
to which water is upheld is El. 703.6, or 14 feet higher. The dam 
actually holds up 31 feet of water above the bed of the river. 



Fig. 121. Lauchli Automatic Sluice Gate 

This system is open to the following objections: First, the 
immense expense involved in a triple row of steel girders of large 
span carrying the curtains and their apparatus; and second, the 
large surface exposure to wind which must always be a menace to 
the safety of the curtains. 

It is believed that the raising of the water level could be effected 
for a quarter of the cost if not much less, by adopting a combination 
of the system used in the Folsam v T eir, Fig. 50, with that in the 
Dhukwa weir, Fig. 52, viz, hinged collapsible gates which could be 
pushed up or lowered by hydraulic jacks as required. The existing 
lower part of the dam could be. utilized and a subway constructed 




















































































































DAMS AND WEIRS 


205 


through it for cross communication and accommodation for the 
pressure pipes, as is the case in the Dhukwa weir. This arrange¬ 
ment which is quite feasible would, it is deemed, be an improvement 
on the expensive, complicated, and slow, Carnere curtain system. 

154. Automatic Dam or Regulator. Mr. Lauchli of New York, 
writing for Engineering News, describes a new design for automatic 
regulators, as follows: 

In Europe there has been in operation for some time a type of automatic 
dam or sluice gate which on account of its simplicity of construction, adapt- 



Fig. 122. View of Lauchli Automatic Dam Which Has Been for Several Years 

in Successful Operation in Europe 

ability to existing structures, exact mathematical treatment, and especially 
its successful operation, deserves to attract the attention of the hydraulic engineer 
connected with the design of hydroelectric plants or irrigation w r orks. Fig. 
121 shows a cross-section and front elevation of one of the above-mentioned 
dams now in course of construction, and the view in Fig. 122 gives an idea of 
a small automatic dam of the same type which has been in successful operation 
for several seasons, including a severe winter, and during high spring floods. 

Briefly stated, the automatic dam is composed of a movable part or 
panel, resting at the bottom on a knife edge, and fastened at the top to a com¬ 
pensating roller made of steel plate and filled with concrete. This roller moves 
along a track located at each of its ends, and is so designed as to take, at any 
height of water upstream, a position such as will give the apron the inclination 
necessary for discharging a known amount of water, and in so doing will keep 
the upper pool at a constant fixed elevation. 

With the roller at its highest position the panel lies horizontally, and 
the full section is then available for discharging water. Any debris, such as 









206 


DAMS AND WEIRS 


trees, or ice cakes, etc., will pass over the dam without any difficulty, even 
during excessive floods, as the compensating roller is located high above extreme 
flood level. 

The dam now in course of construction is located on the river Grafenauer 
Ohe, in Bavaria, and will regulate the water level at the intake of a paper mill, 
located at some distance from the power house. The dam has a panel 24.27 
ft. long, 6.85 ft. high, and during normal water level will discharge 1400 cu. ft. 
per sec., while at flood time it will pass 3,530 cu. ft. per sec. of water. As shown 
in Fig. 121, the main body of the dam is made of a wooden plank construction 
laid on a steel frame. The panel is connected with the compensating roller 
at each end by a flexible steel cable wound around the roller end, and then 
fastened at the upper part of the roller track to an eyebolt. A simple form of 
roof construction protects the roller track from rain and snow. The panel 
is made watertight at each extremity by means of galvanized sheet iron held 
tight against the abutments by water pressure. This type of construction has 
so far proved to be very effective as to watertightness. 

It may be needless to point out that this type of dam can also be fitted 
to the crest of overflow dam of ordinary cross-section, and then fulfill the duty 
of movable dashboards. 

The probability is that this type will become largely used in 
the future. A suggested improvement would be to abolish the cross 
roller having instead separate rollers on each pier or abutment, 
working independently. There will then be no practical limit to 
the span adopted. 


i 


INDEX 




INDEX 


PAGE 

A 

Aprons_ 

decrease uplift, rear aprons_ 71 

fore, base of dam and_ 98 

hearth and anchored_150 

porous fore_173, 175 

rear_159, 161 

riprap to protect_164 

sloping_169 

uplift, affect_ 70 

Arched dams_101 

characteristics_101 

crest width_104 

examples_104 

Barossa_111 

Bear Valley_104 

Burrin Juick subsidiary_112 

Lithgow_112 

Pathfinder_104 

Shoshone_107 

Sweetwater_109 

profiles_103 

correct_ 103 

theoretical and practical_102 

variable radii, with_112 

vertical water loads, support of- 104 

Arrow Rock dam_ 67 

Assiut barrage_192, 194 

Automatic dam or regulator_205 

B 

Barossa dam_111 

Barrages- 182 

Bassano dam_ 146 

Bear Valley dam_- - -194 

Burrin Juick subsidiary dam- 112 

r; •. - c 

Castle wood weir_ 96 

D 

Damietta and Rosetta weirs- 179 

Dams and weirs_ 1 







































INDEX 

PAGE 

Dams and weirs—continued 

arched_ 101 

definition_ 1 

gravity_ 2 

gravity overfall_ 75 

hollow slab buttress_ 136 

multiple arch or hollow arch buttress_ _ _113 

open dams or barrages_ 182 

submerged weirs founded on sand_151 

Dehri weir_ 178 

Dhukwa weir_ 90 

E 

Ellsworth dam_ 136 

F 

Folsam weir_ 85 

G 

Granite Reef weir_ 92 

Gravity dams_ 2 

design_ 4 

analytical method_._18, 34, 43 

broken line profiles, treatment for_ 41 

calculation, method of_ 11 

crest, high and wide_ 13 

crest width_J_ 9 

failure by sliding or shear, security against_ 31 

graphical method_ 16 

Haessler’s method_36, 42 

height, variation of_ 13 

influence lines_ 31 

maximum stress, formulas for_ 27 

elementary profile, application to_ 28 

limiting height by_ 29 

pressure area in inclined back dam, modified equivalent_ 37 

pressure distribution_23, 25, 26 

pressure limit, maximum_ 27 

pressures in figures, actual_ 34 

profile, practical_ 8 

profile, theoretical_ 4 

profiles, curved back_ 39 

rear widening_ 10 

shear and tension, internal_ 30 

stepped polygon_ 37 

vertical component_ 22 

discussion_ 2 

graphical calculations_ 2 

“middle third” and limiting stress_ 3 

pressure of water on wall_ 2 













































INDEX 

PAGE 

Gravity dams—continued 
discussion 

stress limit, compressive_ 4 

examples_ 53 

Arrow Rock_ 67 

Assuan_ 59 

Burrin Juick_ 65 

Cheeseman Lake __ 53 

Cross River and Ashokan_ 65 

New Croton_ 58 

Roosevelt_ 56 

foundations, special_ 69 

aprons affect uplift_ 70 

aprons decrease uplift, rear_ 71 

ice pressure, gravity dam reinforced against_ 73 

rock below gravel_ 72 

“high”_ 43 

base of dam, silt against_ 50 

partial overfall_ 52 

pentagonal profile to be widened_ 47 

pressure, ice_ 51 

toe of dam, filling against_ 50 

Gravity overfall dams or weirs_ 75 

American dams on pervious foundations_ 97 

« 

analytical method_ 88 

base width, approximate_ 77 

characteristics_ 75 

crest width, approximate_ 77 

depth of overfall, calculation of___-_ 82 

examples_ 83 

Castlewood_ 96 

Dhukwa_ 90 

Folsam_ 85 

Granite Reef_ 92 

Mariquina_ 92 

Nira_-_ 95 

“Ogee”_ 85 

St. Maurice River_ 99 

fore apron, base of dam and_ 98 

graphical process_ 78 

hydraulic conditions-- 93 

pressure, moments of_ 81 

water level, pressures affected by- 79 

Guayabal dam_ 141 

H 

Haessler’s method_36, 42 

Hindia barrage_194, 198 

Hollow slab buttress dam-136 















































INDEX 

PAGE 

Hollow slab buttress dam—continued 

description_ 136 

examples_ 146 

Bassano_146 

Ellsworth_136 

Guayabal_ 141 

fore slope, steel in_ 139 

foundation foredeck, pressure on_ 149 

baffles_160 

buttresses_150 

hearth or anchored apron_150 

reinforced concrete, formulas for_ 137 

slab deck compared with arch deck_ 140 

K 

Khanki weir_ 171 

L 

Laguna weir_ 179 

Lithgow dam_ 112 

M 

Mariquina weir_ 92 

Merala weir_ 171 

Mir Alam dam_ 114 

Multiple arch or hollow arch buttress dams_ 113 

arch, crest width of_ 125 

arches, differential_ 126 

design_ 122 

examples_ 

Belubula_118 

Big Bear Valley_ 131 

Mir Alam_ 114 

Ogden_x._ 120 

inclination of arch to vertical_ 119 

pressure_ 

flood_129 

foundation, on_125 

water, reverse_124 

stresses_ 117 

value_ 113 

N 

Narora weir_ 167 

Nira weir__ 95 






































INDEX 

PAGE 

o 

“Ogee” gravity overfall dam_ 85 

Okhla and Madaya weirs_ 177 

Open dams or barrages_ 182 

advantages_200 

American vs. Indian treatment_196 

automatic_205 

Corbett dam, weir sluices of_189 

definition_ 182 

examples_ 

Assiut__192, 194 

Hindia-194, 198 

North Mon_200 

Upper Coleroon_201 

St. Andrew’s Rapids_201 

Camere type_ 203 

heavy construction_ 190 

moments for Hindia barrage_ 198 

piers, thickness of_200 

regulator, Upper Coleroon_201 

river regulators, features of_ 193 

rivers, across_ 192 

spans, length of_ 197 

weir scouring sluice, example of_ 185 

weir scouring sluice on sand_ 190 

P 

Pathfinder dam_ 104 

S 

St. Andrew’s Rapids dam_201 

Shoshone dam_ 107 

Stepped polygon- 37 

Submerged weirs on sand-151 

apron, rear- 159, 161 

computations, simplifying-156 

crest shutters_ 182 

description_ 151 

examples- 

Damietta and Rosetta_179 

Dehri_ 178 

Khanki_^_ 171 

Laguna_ 179 

Merala_171 

Narora_ 167 

Okhla and Madaya_177 

fore aprons, porous- — 173 

core walls, divided by- 175 













































INDEX 

PAGE 

Submerged weirs on sand—continued 

hydraulic flow, laws of_ 152 

percolation_ 

coefficient of_ 153 

beneath dam_ 152 

values of coefficient of_ 155 

vertical obstruction to_ 159 

pervious dam, paradox of a_ 181 

riprap_ 164 

safety, criterion for_ 154 

sloping apron_ 169 

stability, governing factor for_ 153 

Sweetwater dam_ 109 

T 

Tables_1_ 

pier thickness_ 200 

values of L v or talus width_ 163 

U 

Upper Coleroon regulator_201 


IRBA 8 78 


















I 













